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one promoter per

operon. In contrast, human DNA requires one promoter for each gene.

Complexity may explain some of the differences between the DNA content of bacteria and humans.

But an extension of this line of reasoning would lead to the conclusion that frogs are more complex than

humans because frogs have 8 ft of DNA per diploid nucleus, compared to the 6 ft in a human cell. Logic,

or perhaps vanity, suggests that the amount of DNA per cell does not necessarily reflect the complexity of

the organism. One of the features of frog DNA that may explain its length is that frogs have more repetitive

DNA than humans. More than 75% of the frog genome is in the moderately and highly repetitive category,

whereas only about 35% of the human genome is repetitive.

Major differences between prokaryotic and eukaryotic DNA and RNA are summarized in Table 14.2.

LINEs make up about 5% of the human genome. In some patients with hemophilia (a disease in which

blood does not clot normally), a LINE sequence has been inserted into exon 14 of the gene for factor VIII,

a protein of the blood-clotting system. The insertion of the LINE sequence leads to the production of a

nonfunctional protein.

CLINICAL COM M ENTSLisa N. Patients with β+-thalassemia who maintain their hemoglobin levels >6.0 to 7.0 g/dL are

usually classified as having thalassemia intermedia. In the β-thalassemias, the α-chains of adult

hemoglobin A (α2β2) continue to be synthesized at a normal rate. These chains accumulate in the bone

marrow in which the red blood cells are synthesized during the process of erythropoiesis (generation of

red blood cells). The accumulation of α-chains diminishes erythropoiesis, resulting in an anemia.

Individuals who are homozygous for a severe mutation require constant transfusions. Individuals with thalassemia intermedia, such as Lisa N., could have inherited two different defective

alleles, one from each parent. One parent may be a “silent” carrier, with one normal allele and one mildly

affected allele. This parent produces enough functional β-globin so few or no clinical symptoms of

thalassemia appear. (However, they generally have a somewhat decreased amount of hemoglobin,

resulting in microcytic hypochromic red blood cells.) When this parent contributes the mildly defective

allele and the other heterozygous parent contributes a more severely defective allele, thalassemia

intermedia occurs in the child. The child is thus heterozygous for two different defective alleles.

Isabel S. Isabel S. was treated with a multidrug regimen for tuberculosis because the microbes that

cause the disease frequently become resistant to the individual drugs. The current approach in

patients with M. tuberculosis is to initiate antimycobacterial therapy with four agents because the

mycobacteria frequently become resistant to one or more of the individual antitubercular drugs. The same

approach is taken for patients with HIV, with careful attention given to drug interactions. Isabel was

started on isoniazid (INH), rifampin, pyrazinamide, and ethambutol. Isoniazid inhibits the biosynthesis of

mycolic acids, which are important constituents of the mycobacterial cell wall. Isoniazid is often

prescribed with vitamin B6 (pyridoxine) because isoniazid can interfere with the activation of this vitamin

(to pyridoxal phosphate), which can lead to an alteration in normal cellular metabolism and result in a

clinical neuropathy. Rifampin binds to and inhibits bacterial RNA polymerase, which selectively kills the

bacteria that cause the infection. Pyrazinamide, a synthetic analog of nicotinamide, targets the

mycobacterial fatty acid synthase I gene involved in mycolic acid biosynthesis in M. tuberculosis.

Ethambutol blocks arabinosyl transferases that are involved in cell wall biosynthesis.

Just as bacteria can become resistant to drugs, so can HIV. Because of this concern, patients with HIV

are treated with multidrug regimens. Multidrug regimens usually include two nucleoside reverse

transcriptase inhibitors (NRTIs), such as lamivudine (3TC, Epivir) and abacavir, as well as a third agent.

The third agent is usually either a nonnucleoside reverse transcriptase inhibitor (NNRTI), an example of

which is efavirenz; a protease inhibitor (PI), an example of which is indinavir; or an integrase inhibitor.

PIs prevent the HIV polyprotein from being cleaved into its mature products (see “Biochemical

Comments”). The drugs are often combined into one pill to make it easier to take. Isabel S. was started on

efavirenz as her third drug and was counseled on not getting pregnant because this drug is teratogenic.

Catherine T. The toxin α-amanitin is capable of causing irreversible hepatocellular and renal

dysfunction through inhibition of mammalian RNA polymerases. α-Amanitin is particularly

effective at blocking the action of RNA polymerase II. Fortunately, Catherine T.’s toxicity proved mild.

She developed only gastrointestinal symptoms and slight changes in her hepatic and renal function, which

returned to normal within a few weeks. Treatment was primarily supportive, with fluid and electrolyte

replacement for that lost through the gastrointestinal tract. No effective antidote is available for theAmanita phalloides toxin.

Sarah L. SLE is a multisystem disease characterized by inflammation related to the presence of

autoantibodies in the blood. These autoantibodies react with antigens normally found in the nucleus,

cytoplasm, and plasma membrane of the cell. Such “self” antigen–antibody (autoimmune) interactions

initiate an inflammatory cascade that produces the broad symptom profile of multiorgan dysfunction found

in Sarah L.

Pharmacologic therapy for SLE involves anti-inflammatory drugs and immunosuppressive agents. It

can include nonsteroidal anti-inflammatory drugs (NSAIDs), corticosteroids, antimalarials, or

immunosuppressive drugs. Plaquenil is an antimalarial drug used to treat skin and joint symptoms in SLE,

although its exact mechanism of action in these patients is not fully understood. Sarah was placed on such

a drug regimen.

Studies have indicated that a failure to dispose properly of cellular debris, a normal byproduct of cell death, may lead to the induction of autoantibodies directed against chromatin

in patients with SLE. Normal cells have a finite lifetime and are programmed to die (apoptosis)

through a distinct biochemical mechanism. One of the steps in this mechanism is the stepwise

degradation of cellular DNA (and other cellular components). If the normal intracellular

components are exposed to the immune system, autoantibodies against them may be generated. The

enzyme in cells that degrades DNA is deoxyribonuclease I (DNase I), and individuals with SLE

have reduced serum activity levels of DNase I compared with individuals who do not have the

disease. Through an understanding of the molecular mechanism whereby autoantibodies are

generated, it may be possible to develop therapies to combat this disorder. BIOCHEM ICAL COM M ENTS

Production of the Virus that Causes AIDS. AIDS is caused by the HIV. Two forms of the virus

have been discovered, HIV-1, which is prevalent in industrialized countries, and HIV-2, which is

prevalent in certain regions of Africa. Eight to 10 years or more can elapse between the initial infection

and development of the full-blown syndrome.

Proteins in the viral coat bind to membrane protein receptors (named CD4) of helper T-lymphocytes,

a class of cells involved in the immune response. Subsequently, conformational changes occur that allow

the viral-coat proteins to bind to a chemokine coreceptor in the cell membrane. The lipid in the viral coat

then fuses with the cell membrane, and the viral core enters the cell, releasing its RNA and enzymes

(including the reverse transcriptase) by a process called uncoating. Reverse transcriptase uses the viral

RNA as a template to produce a single-stranded DNA copy, which then serves as a template for synthesis

of a double-stranded DNA. An integrase enzyme, also carried by the virus, enables this DNA to integrate

into the host cell genome as a provirus (Fig. 14.22).In the initial stage of transcription of the provirus, the transcript is spliced, and three proteins—Nef, Tat, and Rev—are produced. Tat stimulates transcription of the viral genes. As Rev accumulates, it

allows unspliced viral RNA to leave the nucleus and to produce proteins of the viral envelope and viral

core, including reverse transcriptase. Two of the envelope glycoproteins (gp41 and gp120, which are

derived from the env gene product) form a complex that embeds in the cell membrane. The other proteins,

which are translated as a polyprotein and cleaved by the viral protease (one of the targets of anti-HIV

drugs), combine with the full-length viral RNA to form core viral particles, which bud from the cell

membrane. Thus, the virus obtains its lipid coat from the host cell membrane, and the coat contains the

viral proteins gp41 and gp120. These surface proteins of the virus bind to CD4 receptors on other human

helper T-lymphocytes, and the infection spreads.In an uninfected person, helper T-lymphocytes usually number approximately 1,000/mL. Infection with

HIV causes the number of these cells to decrease, which results in a deficiency of the immune system.

When the number of T-lymphocytes drops to <200/mL, the disease is in an advanced stage, and

opportunistic infections, such as tuberculosis, occur. Although macrophages and dendritic cells lack CD4

receptors, they can also become infected with HIV and can carry the virus to the central nervous system.

The most effective means of combating HIV infection involves the use of drugs that inhibit the viral

reverse transcriptase or the viral protease. However, these drugs only hold the infection at bay; they do

not effect a cure.

Drugs currently used to treat HIV act on the viral reverse transcriptase or the

protease (see

Fig. 14.22). The nonnucleoside drugs (e.g., efavirenz) bind to reverse transcriptase and

inhibit its action. The nucleoside analogs (e.g., lamivudine) add to the 3′-end of the growing DNA

transcript produced by reverse transcriptase and prevent further elongation. The PI (e.g.,

indinavir) bind to the protease and prevent it from cleaving the polyprotein. KEY CONCEPTS

Transcription is the synthesis of RNA from a DNA template.

The enzyme RNA polymerase transcribes genes into a single-stranded RNA.

The RNA produced is complementary to one of the strands of DNA, which is known as the template

strand. The other DNA strand is the coding, or sense, strand.

Bacteria contain a single RNA polymerase; eukaryotic cells use three different RNA polymerases.

The DNA template is copied in the 3′-to-5′ direction and the RNA transcript is synthesized in the 5′-

to-3′ direction.

In contrast to DNA polymerases, RNA polymerases do not require a primer to initiate transcription,

nor do they contain extensive error-checking capabilities.

Promoter regions, specific sequences in DNA, determine where on the DNA template RNA polymerase binds to initiate transcription.

Transcription initiation requires several protein factors to allow for efficient RNA polymerase

binding to the promoter.

Other DNA sequences, such as promoter-proximal elements and enhancers, affect the rate of

transcription initiation through the interactions of DNA-binding proteins with RNA polymerase and

other initiation factors.

Eukaryotic genes contain exons and introns. Exons specify the coding region of proteins, whereas

introns have no coding function.

The primary transcript of eukaryotic genes is modified to remove the introns (splicing) before a

final, mature mRNA is produced.

Table 14.3 summarizes the diseases discussed in this chapter.REVIEW QUESTIONS—CHAPTER 14

1.A gene would need to contain which one of the following templates to generate the short transcript

AUCCGUACG (note that all sequences are written from 5′ to 3)?′ A. ATCCGTACG

B. CGTACGGAT C. AUCCGUACG D. TAGGCATGC E. GCATGCCTA

2.Given that the LD50 (the dose at which 50% of the recipients die) of amanitin is 0.1 mg/kg of body

weight, and that the average mushroom contains 7 mg of amanitin, how many mushrooms must be

consumed by Catherine T. (50 kg of body weight) to be above the LD50? A. 1

B. 2 C. 3 D. 4 E. 5

3.Mutations in DNA large distances from a structural gene can lead to overor underexpression of that

gene. Which one of the following eukaryotic DNA control sequences does not need to be in a fixed

location and is most responsible for high rates of transcription of particular genes?

A. Promoter

B.Promoter-proximal element

C.Enhancer

D.Operator

E.Splice donor site

4.Which one of the following is true of both eukaryotic and prokaryotic gene expression and would

therefore not be an effective target for drug development?A. After transcription, a 3′-poly(A) tail and a 5′-cap are added to mRNA.

B. Translation of mRNA can begin before transcription is complete. C. mRNA is synthesized in the 3′-to-5′ direction.

D. RNA polymerase binds at a promoter region upstream of the gene.

E. Mature mRNA is always precisely collinear to the gene from which it was transcribed.

5.A family has two children, both of whom have a form of β-thalassemia. One child is almost

nonsymptomatic, whereas the other requires frequent blood transfusions for his disease. The α-

globin to β-globin ratio in the more severely affected child is most likely to be which one of the

following? A. 5:1

B. 2:1 C. 1:1 D. 1:2 E. 1:5

6.Certain drugs can be used as antibiotics because they affect bacterial RNA polymerases but not

eukaryotic RNA polymerases. RNA polymerase is a key enzyme in the process of transcription,

which can be best described by which one of the following?

A. The single-stranded RNA produced is identical to one of the strands of double-stranded DNA.

B. The single-stranded RNA produced is identical to both of the strands of double-stranded DNA.

C. Eukaryotic genes are transcribed in the cytosol by three different RNA polymerases.

D. RNA polymerase cannot initiate new strand synthesis and must have a primer. E. Eukaryotic genes are transcribed in the nucleus by three different DNA polymerases.

7.Eukaryotic cells contain multiple RNA polymerases, which makes it difficult to block all RNA

synthesis with one drug targeted to a specific polymerase. Which one of the following best describes

properties of eukaryotic RNA polymerases? A. Polymerase I produces most of the rRNA. B. Polymerase II produces most of the tRNA. C. Polymerase III produces most of the mRNA.

D. All three RNA polymerases have the same mechanism of action and bind to the same promotor

sequences on DNA.

E. Enhancers identify the start point for transcription for all three polymerases.

8.The production of mRNA in eukaryotic cells requires a large number of steps, and drugs targeted to

any of these steps could block mRNA production. Which one of the following accurately describes a

part of the process of producing mRNA from a eukaryotic gene?

A. The sense strand of DNA is the strand used by RNA polymerase during transcription.

B. The antisense DNA strand is identical to the RNA transcript except that the DNA strand contains

thymine and the RNA strand contains uracil.

C. The first RNA form produced contains both intron and exon sequences.

D. Mature mRNA contains a cap at the 5′-end, a poly(A) tail at the 3′ end, introns, and exons.

E. During processing in the nucleus of a precursor mRNA, introns and exons are

A.A specific tRNA can code for multiple different amino acids.

B.tRNA contains a codon site that binds with an anticodon of mRNA.

C.tRNA contains one specific binding site for both the sequence of three nucleotides in mRNA and

the encoded amino acid.

D.One of the loops of tRNA contains the anticodon.

E.The D loop contains the anticodon.

10. A researcher wants to develop an antibiotic that targets histones and introns in bacteria, and she has

applied for a grant. Why would the grant’s physician/biochemist advisor advise against funding this

grant application?

A.The proposed antibiotic would have no effect on bacteria but could harm human cells.

B.The proposed antibiotic would negatively affect both bacteria and human cells.

C.The proposed antibiotic would have no effect on either bacteria or human cells.

D.Bacteria have histones but do not have introns.

E.Bacteria have introns but do not have histones.

ANSWERS TO REVIEW QUESTIONS

1.The answer is B. The transcript that is produced is copied from the DNA template strand, which

must be of the opposite orientation from the transcript. So the 5′-end of the template strand should

base-pair with the 3′-end of the transcript, or the G. Thus, CGTACGGAT would base-pair with

the transcript and would represent the template strand.

2.The answer is A. Catherine T. weighs 50 kg, and if 0.1 mg/kg of body weight is the LD50, then

for Amanda, 5 mg of toxin would bring her to the LD50. Because one mushroom contains 7 mg of

the toxin, ingesting just one mushroom could be fatal.

3.The answer is C. Enhancer sequences can be thousands of bases away from the basal promoter

and still stimulate transcription of the gene. This is accomplished by looping of the DNA so that

the proteins binding to the enhancer sequence (transactivators) can also bind to proteins bound to

the promoter (coactivators). A promoter-proximal element is a DNA sequence near to the

promoter that can bind transcription factors that aid in recruiting RNA polymerase to the promoter

region.

4.The answer is D. Both prokaryotes and eukaryotes require RNA polymerase binding to an

upstream promoter element. Answer A applies only to eukaryotes; prokaryote mRNA is not

capped, nor does it contain a poly(A) tail. Prokaryotes have no nucleus; therefore, the 5′-end of an

mRNA is immediately available for ribosome binding and initiation of translation (thus, B is

incorrect). Answer C is incorrect overall; RNA synthesis, like DNA synthesis, is always in the 5′-

to-3′ direction. Answer E is incorrect because introns are present only in eukaryotic genes.

5.The answer is A. A β-thalassemia refers to a condition in which the α-chain of globin is producedin excess of the β-chain. The greater the ratio of α- to β-chain, the more severe the disease.

6.The answer is A. The single-stranded RNA produced by transcription is identical in sequence to

one (not both) of the DNA strands except that the RNA strand contains the base uracil in locations

where the DNA strand contains thymine. Eukaryotic genes are transcribed in the nucleus (not

cytosol) by three different RNA polymerases (not DNA polymerases). RNA polymerase does not

require a primer to initiate transcription, unlike DNA polymerase, which does require a primer to

initiate DNA replication.

7.The answer is A. The promotor region on the DNA identifies the start point of transcription for

each gene. Enhancers are distal promoter elements that stabilize RNA polymerase binding to the

promoter, but enhancers do not identify the initiation point for transcription. All of the RNA

polymerases have the same mechanism of action but differ in which promoters they recognize (the

sequence of promoters differs for each polymerase) owing to the use of different accessory factors

in forming the initiation complex. Polymerase I produces rRNA; polymerase II, mRNA; and

polymerase III, tRNA in eukaryotic cells.

8.The answer is C. The sense (or coding) strand of DNA is identical to the mRNA produced, with

the exception of the DNA containing thymine and the RNA containing uracil. The template (or

antisense) strand of DNA is the strand that is used by RNA polymerase to produce a complementary RNA sequence to the template strand (thus, the antisense strand is complementary

to the mRNA produced and is not identical to it). The first RNA form produced by RNA polymerase is hnRNA, which contains both intron (noncoding) and exon (coding sequences)

sequences. hnRNA is modified by the addition of a cap to the 5′-end, a poly(A) tail added to the

3′-end, and all introns removed. During processing, the intron sequences are removed from the

hnRNA to produce the mature mRNA.

9.The answer is D. tRNA contains a three-base sequence known as the anticodon, which binds to a

corresponding complementary codon on the mRNA. The amino acid is covalently linked to the

tRNA at its 3′-end, which is distinct from the anticodon site. A particular tRNA only links to one

amino acid, not multiple amino acids. In the cloverleaf structure of tRNA, one of the loops

contains the anticodon that is distinct from the D loop, which frequently contains dihydrouridine as

an unusual base.

10.The answer is A. Bacteria do not have histones or introns, but humans have both. The proposed

antibiotic would have no effect on bacteria but could have a deleterious effect on human cells.15 Translation: Synthesis of Proteins

For additional ancillary materials related to this chapter, please visit thePoint. Proteins are produced by the process of translation, which occurs on ribosomes and is directed by

messenger RNA (mRNA). The genetic message encoded in DNA is first transcribed into mRNA, and the

nucleotide sequence in the coding region of the mRNA is then translated into the amino acid sequence of

the protein.

Translation of the Code. The portion of mRNA that specifies the amino acid sequence of the protein is

read in codons, which are sets of three nucleotides that specify individual amino acids (Fig 15.1). The

codons on mRNA are read sequentially in the 5′-to-3′ direction, starting with the 5′- AUG (or “start”

codon) that specifies methionine and sets the reading frame and ending with a 3′- termination (or

“stop”) codon (UAG, UGA, or UAA). The protein is synthesized from its N terminus to its C terminus.Each amino acid is carried to the ribosome by an aminoacyl–transfer RNA (tRNA) (i.e., a tRNA

with an amino acid covalently attached). Base pairing between the anticodon of the tRNA and the codon

on the mRNA ensures that each amino acid is inserted into the growing polypeptide at the appropriate

position.

Synthesis of the Protein. Initiation involves formation of a complex containing the initial methionyltRNA bound to the AUG “start” codon of the mRNA and to the “P” site of the ribosome. It requires

guanosine triphosphate (GTP) and proteins known as eukaryotic initiation factors (eIFs).

Elongation of the polypeptide involves three steps: (1) binding of an aminoacyl-tRNA to the “A”

site on the ribosome, where it base-pairs with the second codon on the mRNA; (2) formation of a

peptide bond between the first and second amino acids; and (3) translocation, movement of the mRNA

relative to the ribosome, so that the third mRNA codon moves into the “A” site. These three elongation

steps are repeated until a termination codon aligns with the site on the ribosome where the next

aminoacyl-tRNA would normally bind. Release factors bind instead, causing the completed protein to be

released from the ribosome.

After one ribosome binds and moves along the mRNA, translating the polypeptide, another ribosome

can bind and begin translation. The complex of a single mRNA with multiple ribosomes is known as a

polysome.

Folding and Modification and Targeting of the Protein. Folding of the polypeptide into its threedimensional configuration occurs as the polypeptide is being translated. This process involves proteins

called chaperones. Modification of amino acid residues in a protein occurs during or after translation.

Proteins synthesized on cytosolic ribosomes are released into the cytosol or transported into

mitochondria, peroxisomes, and the nucleus. Proteins synthesized on ribosomes attached to the rough

endoplasmic reticulum (RER) are destined for lysosomes, cell membranes, or secretion from the cell.

These proteins are transferred to the Golgi complex, where they are modified and targeted to their

ultimate locations.THE WAITING ROOM

Lisa N., a 4-year-old patient with β+-thalassemia intermedia (see Chapter 14), showed no

improvement in her symptoms at her second visit. Her hemoglobin level was 7.3 g/dL (reference

range for females = 12 to 16 g/dL).

Jay S. is a 9-month-old male infant of Ashkenazi Jewish parentage. His growth and development

were normal until age 5 months, when he began to exhibit mild, generalized muscle weakness. By 7

months, he had poor head control and slowed development of motor skills, and he was increasingly

inattentive to his surroundings. His parents also noted unusual eye movements and staring episodes. On

careful examination of his retinae, his pediatrician observed a “cherry red” spot within a pale macula.

The physician suspected Tay–Sachs disease and sent samples of his whole blood to the molecular

biology–genetics laboratory.

The results of tests performed in the molecular biology laboratory show that Jay S. has an

insertion in exon 11 of the α-chain of the hexosaminidase A gene, the most common mutation

found in patients of Ashkenazi Jewish background who have Tay–Sachs disease. Hexosaminidase

A, the enzyme activity that is lacking in Jay S., can be assayed using a serum sample and a

substrate that releases a fluorescent dye upon being hydrolyzed. When measuring enzyme activity,

one needs to be careful to distinguish between hexosaminidase A activity and a closely related

activity from hexosaminidase B. This is accomplished by differential heat inactivation of the

sample (exposure of the sample to 50°C will inactivate hexosaminidase A activity but not

hexosaminidase B activity). For prenatal screening, molecular techniques are the preferred

method because of their sensitivity and the limited amount of sample available (see Chapter 17).

Paul T. returned to his physician’s office after 5 days of azithromycin therapy (see Chapter 12)

feeling significantly better. The sputum sample from his previous visit had been cultured. The

results confirmed that his respiratory infection was caused by Streptococcus pneumoniae and that the

organism was sensitive to penicillin, macrolides (e.g., erythromycin, clarithromycin), tetracycline, and

other antibiotics.

Edna R., a 25-year-old junior medical student, brings her healthy 4-month-old daughter, Beverly,

to the pediatrician for her second diphtheria, tetanus, and pertussis (DTaP, acellular pertussis)

immunization, along with the following immunizations: pneumococcal, inactivated polio, Haemophilus

influenza, and rotavirus. Edna tells the doctor that her great-great aunt had died of diphtheria during an

epidemic many years ago. I. The Genetic Code

Transcription, the transfer of the genetic message from DNA to RNA, and translation, the transfer of the

genetic message from the nucleotide language of nucleic acids to the amino acid language of proteins, bothdepend on base pairing. In the late 1950s and early 1960s, molecular biologists attempting to decipher the

process of translation recognized two problems. The first involved decoding the relationship between the

language of the nucleic acids and the language of the proteins, and the second involved determining the

molecular mechanism by which translation between these two languages occurs.

Twenty different amino acids are commonly incorporated into proteins, and therefore, the protein

alphabet has 20 characters. The nucleic acid alphabet, however, has only four characters, corresponding

to the four nucleotides of mRNA (A, G, C, and U). If two nucleotides constituted the code for an amino

acid, then only 42 (or 16) amino acids could be specified. Therefore, the number of

nucleotides that code

for an amino acid has to be at least three, providing 43 (or 64) possible combinations or codons—more

than required, but not excessive.

Scientists set out to determine the specific codons for each amino acid. In 1961, Marshall Nirenberg

produced the first crack in the genetic code (the collection of codons that specifies all the amino acids

found in proteins). He showed that poly(U), a polynucleotide in which all the bases are uracil, produced

polyphenylalanine in a cell-free protein-synthesizing system. Thus, UUU must be the codon for

phenylalanine. As a result of experiments using synthetic polynucleotides in place of mRNA, other codons

were identified.

The pioneering molecular biologists recognized that, because amino acids cannot bind directly to the

sets of three nucleotides that form their codons, adapters are required. The adapters were found to be

transfer RNA (tRNA) molecules. Each tRNA molecule contains an anticodon and covalently binds a

specific amino acid at its 3′-end (see Chapters 12 and 14). The anticodon of a tRNA molecule is a set of

three nucleotides that can interact with a codon on mRNA (see Fig. 15.1). In order to interact, the codon

and anticodon must be complementary (i.e., they must be able to form base pairs in an antiparallel

orientation). Thus, the anticodon of a tRNA serves as the link between an mRNA codon and the amino

acid that the codon specifies.

Obviously, each codon present within mRNA must correspond to a specific amino acid. Nirenberg

found that trinucleotides of known base sequence could bind to ribosomes and induce the binding of

specific aminoacyl-tRNAs (i.e., tRNAs with amino acids attached covalently). As a result of these and

the earlier experiments, the relationship between all 64 codons and the amino acids they specify (the

entire genetic code) was determined by the mid-1960s (Table 15.1).Three of the 64 possible codons (UGA, UAG, and UAA) terminate protein synthesis and are known as stop or nonsense codons. The remaining 61 codons specify amino acids. Two amino acids each have only

one codon (AUG for methionine; UGG for tryptophan). The remaining amino acids have multiple codons.

A. The Code Is Degenerate Yet Unambiguous

Because many amino acids are specified by more than one codon, the genetic code is described as

degenerate, which means that an amino acid may have more than one codon. However, each codon

specifies only one amino acid, and the genetic code is thus unambiguous. Inspection of a codon table shows that in most instances of multiple codons for a single amino acid,

the variation occurs in the third base of the codon (see Table 15.1). Crick noted that the pairing between

the 3′-base of the codon and the 5′-base of the anticodon does not always follow the strict base-pairing

rules that he and Watson had previously discovered (i.e., A pairs with U, and G with C). This observation

resulted in the wobble hypothesis.

At the third base of the codon (the 3′-position of the codon and the 5′-position of the anticodon), the

base pairs can wobble. For example, G can pair with U, and A, C, or U can pair with the unusual base

hypoxanthine (I) found in tRNA. Thus, three of the four codons for alanine (GCU, GCC, and GCA) can