CONCEPT MAP: THE GENERATION OF BIOCHEMICAL ENERGY
25.10 Translation: Transfer RNA and Protein Synthesis
How are the messages carried by mRNA translated, and how does the translation pro- cess result in the synthesis of proteins? Protein synthesis occurs at ribosomes, which are located outside the nucleus in the cytoplasm of cells. First, mRNA binds to the ribosome; then, amino acids, which are available in the cytosol, are delivered one by one by transfer RNA (tRNA) molecules to be joined into a specific protein by the ribo- somal “machinery.” All of the RNA molecules required for translation were synthesized from DNA by transcription in the nucleus and moved to the cytosol for translation.
First, let us examine the structure of the tRNAs. Every cell contains more than 20 dif- ferent tRNAs, each designed to carry a specific amino acid, even though they are all simi- lar in overall structure. A tRNA molecule is a single polynucleotide chain held together by regions of base pairing in a partially helical structure something like a cloverleaf (Figure 25.7a). In three dimensions, a tRNA molecule is L-shaped, as shown in Figures 25.7b and c.
At one end of the L-shaped tRNA molecule, an amino acid is bonded to its spe- cific tRNA by an ester linkage between the iCOOH of the amino acid and an
iOH group on the last ribose at the 3⬘ end of the tRNA chain. Individual synthe- tase enzymes are responsible for connecting each amino acid with its partner tRNA in an energy- requiring reaction. This reaction is referred to as charging the tRNA. Once charged, the tRNA is ready to be used in the synthesis of new protein.
▶ Figure 25.7 Structure of tRNA.
(a) Schematic, flattened tRNA mol- ecule. The cloverleaf-shaped tRNA contains an anticodon triplet on one
“leaf ” and a covalently bonded amino acid at its 3⬘ end. The example shown is a yeast tRNA that codes for phe- nylalanine. All tRNAs have similar structures. The nucleotides not identi- fied (blank circles) are slightly altered analogs of the four normal ribonu- cleotides. (b) A computer-generated model of the serine tRNA molecule.
The serine binding site is shown in yel- low and the anticodon in red. (c) The three-dimensional shape (the tertiary structure) of a tRNA molecule. Note how the anticodon is at one end and the amino acid is at the other end.
DNA
mRNA leaves nucleus Transcription
Translation (at ribosome)
Cell nucleus
Cytoplasm mRNA
Polypeptide
tRNAs
▲ Overview of protein synthesis.
The codons of mature mRNA are translated in the ribosomes, where tRNAs deliver amino acids to be assembled into proteins (polypeptides).
S E C T I O N 2 5 . 1 0 Translation: Transfer RNA and Protein Synthesis 797 At the other end of the tRNA “L” is a sequence of three nucleotides called an anticodon
(Figure 25.7). The anticodon of each tRNA is complementary to an mRNA codon—always the one designating the particular amino acid that the tRNA carries. For example, the tRNA carrying the amino acid leucine, which is coded for by 5⬘ CUG 3⬘ in mRNA, has the com- plementary sequence 3⬘ GAC 5⬘ as its anticodon on the tRNA. This is how the genetic message of nucleotide triplets, the codons, is translated into the sequence of amino acids in a protein. When the tRNA anticodon pairs off with its complementary mRNA codon, leucine is delivered to its proper place in the growing protein chain.
The three stages in protein synthesis are initiation, elongation, and termination. As you read the descriptions, follow along in the diagram of translation in Figure 25.8.
Translation Initiation
Each ribosome in a cell is made up of two subunits of markedly different sizes, called, logically enough, the small subunit and the large subunit. Each subunit contains protein enzymes and ribosomal RNA (rRNA). Protein synthesis begins with the binding of an mRNA to the small subunit of a ribosome, joined by the first tRNA. The first codon on the 5⬘ end of mRNA, an AUG, acts as a “start” signal for the translation machin- ery and codes for a methionine-carrying tRNA. Initiation is completed when the large ribosomal subunit joins the small one and the methionine-bearing tRNA occupies one of the two binding sites on the united ribosome. (Not all proteins have methionine at one end. If it is not needed, the methionine from chain initiation is removed by post- translational modification before the new protein goes to work.)
Translation Elongation
Next to the first binding site on the ribosome is a second binding site where the next codon on mRNA is exposed and the tRNA carrying the next amino acid will be attached. All available tRNA molecules can approach and try to fit, but only one with the appropriate anticodon sequence can bind. Once the tRNA with amino acid 2 arrives, a ribozyme in the large subunit catalyzes formation of the new peptide bond and breaks the bond linking amino acid 1 to its tRNA. These energy-requiring steps are fueled by the hydrolysis of GTP to GDP. The first tRNA then leaves the ribosome, and the entire ribosome shifts one codon (three positions) along the mRNA chain. As a result, the second binding site is opened up to accept the tRNA carrying the next amino acid.
The three elongation steps now repeat:
• The next appropriate tRNA binds to the ribosome.
• Peptide bond formation attaches the newly arrived amino acid to the growing chain and the tRNA carrying it is released.
• Ribosome position shifts to free the second binding site for the next tRNA.
A single mRNA can be “read” simultaneously by many ribosomes. The growing polypeptides increase in length as the ribosomes move down the mRNA strand.
Translation Termination
Completed polypeptide
Ribosomal subunits
When synthesis of the protein is completed, a “stop” codon signals the end of transla- tion. An enzyme called a releasing factor then catalyzes cleavage of the polypeptide chain from the last tRNA; the tRNA and mRNA molecules are released from the ribo- some, and the two ribosome subunits separate. This step also requires energy from GTP. Overall, to add one amino acid to the growing polypeptide chain requires 4 mol- ecules of GTP, excluding the energy needed to charge the tRNA.
Anticodon A sequence of three ribo- nucleotides on tRNA that recognizes the complementary sequence (the codon) on mRNA.
In our discussion in this and the preceding sections, we have left many questions about replication, transcription, and translation unanswered. What tells a cell when to start replication? Are there mechanisms to repair damaged DNA or correct random errors made during replication? (There are.) Keep in mind that synthesis of mRNA is the beginning of synthesis of a protein. How are proteins modified, and where does this modification occur? Since each cell contains the entire genome and since cells differ widely in their function, what keeps genes for unneeded proteins from being
Adenine Cytosine Guanine Uracil (RNA)
INITIATION begins with small ribosomal subunit and the first tRNA arriving at the start codon of the mRNA.
The small and large ribosomal units interlock around the mRNA, with the first tRNA in place at the start codon, completing the initiation stage. The tRNA with amino acid 2 is approaching.
Nucleus
mRNA Small
ribosomal subunit
Start codon
mRNA strand tRNA binding sites
Anticodon tRNA Amino acid
ELONGATION begins as the tRNA with amino acid 2 binds to its codon at the second site within the ribosome.
A peptide bond forms between amino acid 1 and 2, the first tRNA is released, the ribosome moves one codon to the right, and the tRNA with amino acid 3 is arriving.
Elongation continues with three amino acids in the growing chain and the fourth one arriving with its tRNA.
TERMINATION occurs after the elongation steps have been repeated until the stop codon is reached. The ribosomal units, the mRNA, and the polypeptide separate.
1 1
2
1 2
1 2
3 3
1 2
1 2 3
4 4
5 Completed polypeptide
Large ribosomal subunit Small ribosomal subunit
Multiple elongation steps
Large ribosomal subunit
Peptide bond 3′
3′
3′
3′
5′
5′
5′
5′
3′
3′ 5′
5′
▲ Figure 25.8
Translation: The initiation, elongation, and termination stages in protein synthesis.
S E C T I O N 2 5 . 1 0 Translation: Transfer RNA and Protein Synthesis 799 transcribed? What determines just when a particular gene in a particular cell is tran-
scribed? What indicates the spot on DNA where transcription should begin? How is hnRNA converted into mRNA? How are transcription and the resulting protein syn- thesis regulated? You will see that steroid hormones function by directly entering the nucleus to activate enzyme synthesis (Section 28.5). This is just one mode of gene regulation. It can also occur by modification of DNA during transcription and dur- ing translation. Much is known beyond what we have covered (and if you are curious, take a genetics course!). There are also many questions that we cannot yet fully answer because scientists do not know the answers yet, although research is continuing.
PROBLEM 25.20
What amino acid sequence is coded for by the mRNA base sequence CUC-AUU-CCA-UGC-GAC-GUA?
PROBLEM 25.21
What anticodon sequences of tRNAs match the mRNA codons in Problem 25.20?
PROBLEM 25.22
Why is it difficult to develop a universal influenza vaccine? (See Chemistry in Action: Influenza—Variations on a Theme.)
Influenza—Variations on a Theme
When we talk of “the flu,” we take for granted that it is solely a human condition. Flu is caused by the influenza virus, of which there are three major types—A, B, and C—with many subtypes of each one. Influenza A and B viruses cause human flu epi- demics almost every winter. In the United States, these seasonal epidemics can cause illness in 10% to 20% of the human popu- lation and are associated with an average of 36,000 deaths and 114,000 hospitalizations per year.
Getting a flu shot can prevent illness from types A and B in- fluenza. Influenza type C infections cause a mild respiratory ill- ness and are not thought to cause epidemics. Flu shots do not protect against type C influenza. Unfortunately one shot does not protect you from influenza for life; you have to be re-immunized yearly because the influenza virus mutates rapidly, especially the protein coat. Since influenza viruses are ubiquitous, flu can cause either an epidemic or a pandemic. A disease that quickly and se- verely affects a large number of people and then subsides is an epidemic. A pandemic is a widespread epidemic that may affect entire continents or even the world. Both have occurred.
Can animals get the flu? The answer is yes. Many subtypes of influenza A viruses are also found in a variety of animals, including ducks, chickens, pigs, whales, horses, and seals. Certain strains (subtypes) of influenza A virus are specific only to certain species. Unlike other animals, however, birds are susceptible to all known subtypes of the influenza A virus and serve as reservoirs.
Influenza viruses that infect birds are called avian influenza viruses; first identified in Italy more than 100 years ago, these viruses occur naturally among birds worldwide. Wild birds,
most notably migratory waterfowl such as wild ducks, carry the viruses in their intestines. Avian influenza is very contagious among birds; If infection does occur, domesticated birds, such as chickens, ducks, and turkeys, are particularly susceptible to infection, which either makes them very sick or kills them.
Humans also are susceptible to influenza A viruses, but avian influenza viruses do not usually infect humans due to subtype dif- ferences. However, several cases of human infection with avian influenza viruses have occurred since 1997. These viruses may be transmitted to humans directly from birds, from an environment contaminated by avian virus, or through an intermediate host, such as a pig. Because pigs are susceptible to infection by both avian and human viruses, they can serve as a “mixing vessel” for the scram- bling of genetic material from human and avian viruses, resulting
CHEMISTRY IN ACTION
▲ A tranmission electron micrograph of negatively stained influenza A virus particles.
in the emergence of a novel viral subtype. For example, if a pig is infected with a human influenza virus and an avian influenza virus at the same time, the viruses can re-assort genes and produce a new virus that has most of the genes from the human virus, but surface proteins from the avian virus. This process is known as an antigenic shift. This is how a new virus is formed, a virus against which humans will have little or no immunity and that may result in sustained human-to-human transmission and ultimately an influenza epidemic. Conditions favorable for the emergence of antigenic shift have long been thought to involve humans living in close proximity to domestic poultry and pigs. However, recent events suggest humans themselves can serve as the “mixing ves- sel.” This scenario has frightening consequences; so frightening that the Centers for Disease Control and Prevention (CDC) consid- ers the control of avian influenza to be a top priority. Luckily, the bird flu outbreak in 2006 was limited, although serious, and the swine flu pandemic of 2009 was not as virulent a strain as was first thought. That particular influenza A virus mixed genes from human, avian, and swine viruses, resulting in a novel virus to which humans had no immunity. The 2009 viral strain has some genetic similarities to an older influenza virus that had an alarmingly high mortality rate.
In 1918, a strain of influenza that became known as Spanish flu killed an estimated 20–50 million people worldwide. Most of those who died were not in the highly fragile groups of very young and very old but rather were healthy adults between 20 and 40 years old. Analysis and reconstruction of the 1918 influenza virus showed it was Type A, variant H1N1. The avian influenza virus we are seeing today is also a type A virus, but a different variant than the virus that caused the 1918 pandemic.
However, research has shown that the hemagglutinin genes and gene products (protein) of the avian flu virus are closely simi- lar to those of the 1918 flu virus, suggesting an explanation for why the avian flu is able to infect humans.
Scientists continue to monitor viral sub-strain shifts and drug companies prepare seasonal influenza vaccine based on predictions of what the next season’s predominant viral strains will be. Work is also moving forward on a universal vaccine so that you can avoid the yearly shot and be immunized against influenza like you are against other common viral diseases such as measles.
See Chemistry in Action Problems 25.75 and 25.76 at the end of the chapter.
SUMMARY: REVISITING THE CHAPTER GOALS
1. What are the compositions of the nucleic acids, DNA and RNA? Nucleic acids are polymers of nucleotides. Each nucle- otide contains a sugar, a base, and a phosphate group. The sugar is D-ribose in ribonucleic acids (RNAs) and 2-deoxy-D-ribose in deoxyribonucleic acids (DNAs). The C5iOH of the sugar is bonded to the phosphate group, and the anomeric carbon of the sugar is connected by an N-glycosidic bond to one of five heterocyclic nitrogen bases (Table 25.1). A nucleoside contains a sugar and a base, but not the phosphate group. In DNA and RNA, the nucleotides are connected by phosphate diester link- ages between the 3⬘iOH group of one nucleotide and the 5⬘ phosphate group of the next nucleotide. DNA and RNA both contain adenine, guanine, and cytosine; thymine occurs in DNA and uracil occurs in RNA (see Problems 1–5, 23, 29–32, 47–50).
2. What is the structure of DNA? The DNA in each chromo- some consists of two polynucleotide strands twisted together in a double helix. The sugar–phosphate backbones are on the out- side, and the bases are in the center of the helix. The bases on the two strands are complementary—opposite every thymine is an adenine, opposite every guanine is a cytosine. The base pairs are connected by hydrogen bonds (two between T and A; three between G and C). Because of the base pairing, the DNA strands are antiparallel: One DNA strand runs in the 5⬘ to 3⬘ direction and its complementary partner runs in the 3⬘ to 5⬘ direction (see Problems 6–11, 24, 26, 37–44, 51, 52, 72, 81).
3. How is DNA reproduced? Replication (Figure 25.5) requires DNA polymerases and deoxyribonucleoside triphosphates. The DNA helix partially unwinds and the enzymes move along the separated DNA strands, synthesizing a new strand with bases
complementary to those on the unwound DNA strand being cop- ied. The enzymes move only in the 3⬘ to 5⬘ direction along the template strand (and thus new DNA strands only grow in the 5⬘ to 3⬘ direction), so that one strand is copied continuously and the other strand is copied in segments as the replication fork moves along. In each resulting double helix, one strand is the original template strand and the other is the new copy (see Problems 11, 22, 25, 28, 35, 36, 45, 46, 54, 55, 71).
4. What are the functions of RNA? Messenger RNA (mRNA) carries the genetic information out of the nucleus to the ribo- somes in the cytosol, where protein synthesis occurs. Transfer RNAs (tRNAs) circulate in the cytosol, where they bond to amino acids that they then deliver to ribosomes for protein synthesis.
Ribosomal RNAs (rRNAs) are incorporated into ribosomes (see Problems 13, 14, 33, 34, 71–74).
5. How do organisms synthesize messenger RNA? In tran- scription (Figure 25.6), one DNA strand serves as the template and the other, the informational strand, is not copied. Nucleo- tides carrying bases complementary to the template bases between a control segment and a termination sequence are connected one by one to form mRNA. The primary transcript mRNA (or hnRNA) is identical to the matching segment of the informational strand, but with uracil replacing thymine. Introns, which are base sequences that do not code for amino acids in the protein, are cut out before the final transcript mRNA leaves the nucleus (see Problems 15–17, 27, 53, 75, 76).
6. How does RNA participate in protein synthesis? The genetic information is read as a sequence of codons—triplets of bases in DNA that give the sequence of amino acids in a protein.
Understanding Key Concepts 801 Of the 64 possible codons (Table 25.4), 61 specify amino acids
and 3 are stop codons. Each tRNA has at one end an anticodon consisting of three bases complementary to those of the mRNA codon that specifies the amino acid it carries. Initiation of transla- tion (Figure 25.8) is the coming together of the large and small subunits of the ribosome, an mRNA, and the first amino acid–
bearing tRNA connected at the first of the two binding sites in
the ribosome. Elongation proceeds as the next tRNA arrives at the second binding site, its amino acid is bonded to the first one, the first tRNA leaves, and the ribosome moves along so that once again there is a vacant second site. These steps repeat until the stop codon is reached. The termination step consists of separa- tion of the two ribosome subunits, the mRNA, and the protein (see Problems 19–21, 56–70, 77–80).
KEY WORDS
Anticodon, p. 797 Base pairing, p. 784 Chromosome, p. 775 Codon, p. 793
Deoxyribonucleotide, p. 779
DNA (deoxyribonucleic acid), p. 775 Double helix, p. 783
Exon, p. 792 Gene, p. 776
Genetic code, p. 793 Genome, p. 788
Heterogeneous nuclear RNA (hnRNA), p. 792
Intron, p. 792
Messenger RNA (mRNA), p. 790 Nucleic acid, p. 776
Nucleoside, p. 777 Nucleotide, p. 776
Replication, p. 786 Ribonucleotide, p. 779 Ribosome, p. 790
Ribosomal RNA (rRNA), p. 790 Ribozyme, p. 790
RNA (ribonucleic acid), p. 776 Transcription, p. 786
Transfer RNA (tRNA), p. 790 Translation, p. 786
UNDERSTANDING KEY CONCEPTS
25.23 Combine the structures below to create a ribonucleotide.
Show where water is removed to form an N-glycosidic linkage and where water is removed to form a phosphate ester. Draw the resulting ribonucleotide structure, and name it.
OH HOCH2
OH OH O
NH2 N H
N H
N O
N O
O−
−O P OH
25.24 Copy the following diagram and use dotted lines to in- dicate where hydrogen bonding occurs between the complemen- tary strands of DNA. What is the sequence of each strand of DNA drawn (remember that the sequence is written from the 5⬘ to 3⬘ end)?
Sugar Phosphate
3′
3′
5′
5′
P S
S P
S P
S
S
S S
P
P
P P S
P
H
N H
N N N N
H N H
N
O N
H
N H N
O N
H H
O N
N N N H N H N CH3
O O
N
N N
N N H
N H
N H
H O
H N
N N N
H CH3
N O
O N
25.25 Copy this simplified drawing of a DNA replication fork:
5′
5′
3′
3′
B C
A