Chapter Fifteen: The Genetic Code and Translation
COMPREHENSION QUESTIONS
1.
What is the one gene, one enzyme hypothesis? Why was this hypothesis an important
advance in our understanding of genetics?
The one gene, one enzyme hypothesis proposed by Beadle and Tatum states that each
gene encodes a single, separate protein. Now that we know more about the nature of
enzymes and genes, it has been modified to the one gene, one polypeptide hypothesis
since many enzymes consist of multiple polypeptides. The original hypothesis helped
establish a linear link between genes (DNA) and proteins.
2.
What three different methods were used to help break the genetic code? What did each
reveal and what were the advantages and disadvantages of each?
Marshall Nirenberg and Johann Heinrich Matthaei used the enzyme polynucleotide
kinase to create homopolymers of synthetic RNAs. Using a cell-free protein
synthesizing system they were able to determine the amino acid coded by each
homopolymer. By this method, the meanings for the amino acids specified by the
codons UUU, AAA, CCC, and GGG were determined. The disadvantage is that the
meanings for only four codons could be determined. The same system was also used to
create copolymers that contained random mixtures of two nucleotides in a known ratio.
Different amino acids in the protein depended on the ratio of the two nucleotides. To
determine or predict the composition of the codons, the frequency of amino acids
produced using the copolymer was compared with the theoretical frequencies expected
for the codons. A disadvantage of this procedure is that it depended on random
incorporation of the nucleotides, which did not always happen. A further problem was
that the base sequence of the codon could not be determined—only the bases contained
within the codon. The redundancy of the code also provided difficulties since several
different codons could specify the same amino acid.
To solve these problems, Nirenberg and Leder mixed ribosomes bound to short
RNAs of known sequences with charged tRNAs. The mixture was passed through a
nitrocellulose filter to which the tRNAs paired to ribosome-mRNA stuck. They next
determined the amino acids attached to the bound tRNAs. Over 50 codons were
identified by this method. The difficulty is that not all tRNAs and codons could be
identified with this method.
Gobind Khorana and his colleagues used a third method. They synthesized RNA
molecules of known repeating sequences. Using a cell-free protein synthesizing system
they produced proteins of alternating amino acids. However, this procedure could not
specify which codon encodes which amino acid.
3.
What are isoaccepting tRNAs?
Isoaccepting tRNAs are tRNA molecules that have different anticodon sequences but
accept the same amino acids.
Chapter Fifteen: The Genetic Code and Translation 187
4.
What is the significance of the fact that many synonymous codons differ only in the
third nucleotide position?
Synonymous codons code for the same amino acid, or, in other words, have the same
meaning. A nucleotide at the third position of a codon pairs with a nucleotide in the
first position of the anticodon. Unlike the other nucleotide positions involved in the
codon-anticodon pairing, this pairing is often weak or “wobbles,” and nonstandard
pairings can occur. Since many synonymous codons differ at only the third nucleotide
position, it is likely that in these codons the “wobble” and nonstandard base-pairing
with the anticodons will result in the correct amino acid being inserted in the protein
even if a nonstandard pairing occurs.
5.
Define the following terms as they apply to the genetic code:
(a) Reading frame
The reading frame refers to each different way that the groups of three nucleotides
or codons can be read in a sequence of nucleotides. For any sequence of
nucleotides, there are potentially three sets of codons that could specify the amino
acid sequence of a polypeptide.
(b) Overlapping code
If an overlapping code is present, then a single nucleotide is included in more than
one codon. The result for a sequence of nucleotides is that more than one type of
polypeptide can be encoded within that sequence.
(c) Nonoverlapping code
In a nonoverlapping code, a single nucleotide is part of only one codon. For a
sequence of RNA, this results in the production of a single type of polypeptide.
(d) Initiation codon
An initiation codon establishes the appropriate reading frame and specifies the first
amino acid of the protein chain. Typically the initiation codon is AUG; however,
both GUG and UUG can also serve as initiation codons.
(e) Termination codon
The termination codon signals the termination or end of translation and the end of
the protein molecule. There are three types of termination codons—UAA, UAG, and
UGA—which can also be referred to as stop codons or nonsense codons. These
codons do not code for amino acids.
(f) Sense codon
A sense codon is a group of three nucleotides that code for an amino acid. There
are 61 sense codons that code for the 20 amino acids commonly found in proteins.
(g) Nonsense codon
Nonsense codons or termination codons signal the end of translation. These codons
do not code for amino acids.
(h) Universal code
In a universal code, each codon specifies or codes for the same amino acid in all
organisms. The genetic code is nearly universal but not completely. Most of the
exceptions occur in mitochondrial genes.
(i) Nonuniversal codons
Most codons are universal (or nearly universal) in that they specify the same amino
acids in almost all organisms. However, there are exceptions where a codon has
188 Chapter Fifteen: The Genetic Code and Translation
different meanings in different organisms. Most of the known exceptions are the
termination codons, which in some organisms do code for amino acids.
Occasionally a sense codon is substituted for another sense codon.
6.
How is the reading frame of a nucleotide sequence set?
The initiation codon on the mRNA sets the reading frame.
7.
How are tRNAs linked to their corresponding amino acids?
For each of the 20 different amino acids that are commonly found in proteins, there is a
corresponding aminoacyl-tRNA synthetase that covalently links the amino acid to the
tRNA molecule.
8.
What role do the initiation factors play in protein synthesis?
Initiation factors are proteins that are required for the initiation of translation. In
bacteria, there are three initiation factors (IF1, IF2, and IF3). Each one has a different
role. IF1 promotes the disassociation of the large and small ribosomal subunits. IF3
binds to the small ribosomal subunit and prevents it from associating with the large
ribosomal subunit. IF2 is responsible for binding GTP and delivering the fMet-tRNAfmet
to the initiator codon on the mRNA. In eukaryotes, there are more initiation factors, but
many have similar roles. Some of the eukaryotic initiation factors are necessary for
recognition of the 5' cap on the mRNA. Others possess a RNA helicase activity, which is
necessary to resolve secondary structures.
9.
How does the process of initiation differ in bacterial and eukaryotic cells?
Bacterial initiation of translation requires that sequences in the 16S rRNA of the small
ribosomal subunit bind to the mRNA at the ribosome binding site or Shine-Dalgarno
sequence. The Shine-Dalgarno sequence is essential in placing the ribosome over the
start codon (typically AUG). In eukaryotes, there is no Shine-Dalgarno sequence. The
small ribosomal subunit recognizes the 5' cap of the eukaryotic mRNA with the
assistance of initiation factors. Next, the ribosomal small subunit migrates along the
mRNA scanning for the AUG start codon. In eukaryotes, the start codon is located with
a consensus sequence called the Kozak sequence (5'–ACCAUGG–3'). Transcription in
eukaryotes also requires more initiation factors.
10.
Give the elongation factors used in bacterial translation and explain the role played by
each factor in translation.
Three elongation factors have been identified in bacteria: EF-TU, EF-TS, and EF-G.
EF-TU joins with GTP followed by a tRNA charged with an amino acid. The charged
tRNA is delivered to the ribosome at the “A” site. During the process of delivery, the
GTP joined to EF-TU is cleaved to form a EF-TU-GDP complex. EF-TS is necessary to
regenerate EF-TU-GTP. The elongation factor EF-G binds GTP and is necessary for
the translocation or movement of the ribosome along the mRNA during translation.
11.
What events bring about the termination of translation?
The process of termination begins when a ribosome encounters a termination codon.
Since the termination codon would be located at the “A” site, no corresponding tRNA
Chapter Fifteen: The Genetic Code and Translation 189
will enter the ribosome. This allows for the release factors (RF1, RF2, and RF3) to bind
the ribosome. RF1 recognizes and interacts with the stop codons UAA and UAG, while
RF2 can interact with UAA and UGA. A RF3-GTP complex binds to the ribosome.
Termination of protein synthesis is complete when the polypeptide chain is cleaved
from the tRNA located at the “P” site. During this process, the GTP is hydrolyzed to
GDP.
12. Give several examples of RNA-RNA interactions that take place in protein synthesis.
Several RNA-RNA interactions that take place during protein synthesis are important.
The tRNA molecules form base pairs with codons on the mRNA. The 3' end of the 16S
rRNA within the small ribosomal subunit forms base pairs with Shine-Dalgarno
sequence at the 5' end of the mRNA. Ribosomal RNAs on both the large and small
subunit interact with tRNAs at both the “A” and the “P” sites. The association of the
large and small subunits of the ribosome is potentially the result of interactions
between the 16S rRNA of the small subunit and the 23S rRNA of the large subunit.
13. What are some types of posttranslational modification of proteins?
Several different modifications can occur to a protein following translation. Frequently
the amino terminal methionine may be removed. Sometimes in bacteria only the formyl
group is cleaved from the N-formyl methionine, leaving a methionine at the amino
terminal. More extensive modification occurs in some proteins that are originally
synthesized as precursor proteins. These precursor proteins are cleaved and trimmed
by protease enzymes to produce a functional protein. Glycoproteins are produced by
the attachment of carbohydrates to newly synthesized proteins. Molecular chaperones
are needed by many proteins to ensure that the proteins are folded correctly. Secreted
proteins that are targeted for the membrane or other cellular locations frequently have
15 to 30 amino acids, called the signal sequence, removed from the amino terminal.
Finally, acetylation of amino acids in the amino terminal of some eukaryotic proteins
also occurs.
14. Explain how some antibiotics work by affecting the process of protein synthesis.
A number of antibiotics bind the ribosome and inhibit protein synthesis at different
steps in translation. Some antibiotics such as streptomycin bind to the small subunit
and inhibit translation initiation. Other antibiotics such as chloramphenicol bind to the
large subunit and block elongation of the peptide by preventing peptide bond
formation. Antibiotics such as tetracycline and neomycin bind the ribosome near the
“A”’ site yet have different effects. Tetracyclines block entry of charged tRNAs to the
“A” site, while neomycin induces translational errors. Finally, some antibiotics such as
erythomycin block the translocation of the ribosome along the mRNA.
15. Compare and contrast the process of protein synthesis in bacterial and eukaryotic cells,
giving similarities and differences in the process of translation in these two types of
cells.
Bacterial and eukaryotic cells share several similarities as well as have several
differences in protein synthesis. Initially bacteria and eukaryotes share the universal
genetic code. However, the initiation codon, AUG, in eukaryotic cells codes for
190 Chapter Fifteen: The Genetic Code and Translation
methionine, whereas in bacteria the AUG codon codes for N-formyl methionine. In
eukaryotes, transcription takes place within the nucleus, whereas most translation takes
place in the cytoplasm (although some translation does take place within the nucleus).
So, transcription and translation in eukaryotes are kept temporally and spatially
separate. However, in bacterial cells transcription and translation occur nearly
simultaneously.
Stability of mRNA in eukaryotic cells and bacterial cells is also different. Bacterial
mRNA is typically short lived, lasting only a few minutes. Eukaryotic mRNA may last
hours or even days. Charging of the tRNAs with amino acids is essentially the same in
both bacteria and eukaryotes. The ribosomes of bacteria and eukaryotes are different
as well. Both bacteria and eukaryotes have large and small ribosomal subunits, but
they differ in size and composition. The bacterial large ribosomal consists of two
ribosomal RNAs, while the eukaryotic large ribosomal subunit consists of three.
During translation initiation, the bacterial small ribosomal subunit recognizes the
Shine-Dalgarno consensus sequence in the 5' UTR of the mRNA and to regions of the
16S rRNA. In most eukaryotic mRNAs, the small subunit binds the 5' cap of the mRNA
and scans downstream until it encounters the first AUG codon. Finally, elongation and
termination in bacterial and eukaryotic cells are functionally similar, although different
elongation and termination factors are used.
APPLICATION QUESTIONS AND PROBLEMS
16.
Sydney Brenner isolated Salmonella typhimurium mutants that were implicated in the
biosynthesis of tryptophan and would not grow on minimal medium. When these
mutants were tested on minimal medium to which one of four compounds (indole
glycerol phosphate, indole, anthranilic acid, and tryptophan) had been added, the
growth responses shown in the table on the facing page were obtained.
Give the order of indole glycerol phosphate, indole, anthranilic acid, and tryptophan
in a biochemical pathway leading to the synthesis of tryptophan. Indicate which step in
the pathway is affected by each of the mutations.
Chapter Fifteen: The Genetic Code and Translation 191
Mutant
Minimal Anthranilic
medium
acid
Indole glycerol
phosphate
Indole
Tryptophan
trp-1
–
–
–
–
+
trp-2
–
–
+
+
+
trp-3
–
–
–
+
+
trp-4
–
–
+
+
+
trp-6
–
–
–
–
+
trp-7
–
–
–
–
+
trp-8
-
+
+
+
+
trp-9
–
–
–
–
+
trp-10
–
–
–
–
+
trp-11
–
–
–
–
+
Based on the mutant strain’s ability to grow on the above substrates, we can group the
mutations into 4 groups that we will call group 1, group 2, group 3, and group 4.
Group 1 mutants can only grow on the minimal medium supplemented with
trpytophan. Group 1: trp 1, trp 10, trp 11, trp 9, trp 6, and trp 7.
Group 2 mutants can grow on the minimal medium supplemented with either
trpytophan or indole. Group 2: trp 3.
Group 3 mutants can grow on the minimal medium supplemented with tryptophan,
indole, or indole glycerol phosphate. Group 3: trp 2 and trp 4.
Group 4 mutants can grow on minimal medium supplemented with the addition of
tryptophan, indole, indole glycerol phosphate, or anthranilic acid. Group 4: trp 8.
By examining the compounds needed for growth by the different groups of mutants,
we can identify the step in the pathway that is blocked by each mutation. For each
group, the pathway step blocked will correspond to the step proceeding the last
compound on which a mutant strain can grow. Any compound added to the minimal
media that proceeds the block will not allow for growth of the mutant strain.
Group 1 mutants can only grow when tryptophan is added to the growth medium.
So group 1 mutants are blocked at the last step in the biosynthesis process before
tryptophan is synthesized. Since group 2 mutants can grow with either tryptophan or
indole added to the growth medium, this suggests that in the pathway indole is the
immediate precursor of tryptophan and that group 2 mutants are blocked in the step
proceeding the synthesis of indole. Using the same type of analysis, we can create the
pathway on the following page for synthesis of tryptophan.
192 Chapter Fifteen: The Genetic Code and Translation
Group 4
Group 3
Group 2 Group 1
Precursor  anthranilic  Indole glyerol  Indole Tryptophan
acid
phosphate
17.
The addition of a series of compounds yielded the following biochemical pathway:
precursor  compound I  compound II  compound III
enzyme A
enzyme B
enzyme C
Mutation a inactivates enzyme A, mutation b inactivates enzyme B, and mutation c
inactivates enzyme C. Mutants, each having one of these defects, were tested on
minimal medium to which compound I, II, and III was added. Fill in the results
expected of these tests by placing a plus sign (+) for growth or a minus sign (–) for no
growth in the following table:
Minimal medium to which is added
Strain with
mutation
Compound I Compound II
a
+
+
b
–
+
c
–
–
Compound III
+
+
+
To determine whether growth will occur on the minimal medium with the added
compound, the step in the pathway where the mutation occurs must be considered.
Since mutation a affects enzyme A, then any strain with mutation a can only grow on
minimal media with the addition of compound I, compound II or compound III. Strains
containing mutation b can only grow with the addition of either compound II or
compound III because enzyme B, which converts compound I to compound II, has been
affected. Strains with mutation c can only grow with the addition of compound III since
the enzyme needed to synthesize compound III from compound II has been mutated.
18.
Assume that the number of different types of bases in RNA is four. What would be the
minimum codon size (number of nucleotides) required if the number of different types
of amino acids in proteins were:
(The number of codons possible must be equal to or greater than the number of
different types of amino acids since the codons encode for the different amino acids. To
calculate how many possible codons that are possible for a given codon size with four
different types of bases in the RNA, the following formula can be used: 4n, where n is
the number of nucleotides within the codon.)
(a) 2
1, because 41 = 4 codons, which is more than enough to specify 2 different amino
acids.
(b) 8
2.
(c) 17
3.
Chapter Fifteen: The Genetic Code and Translation 193
(d) 45
3.
(e) 75
4.
19. How many codons would be possible in a triplet code if only three bases (A, C, and U)
were used?
To calculate the number of possible codons of a triplet code if only three bases are
used, the following equation can be used: 3n, where n is the number of nucleotides
within the codon. So, the number of possible codons is equal to 33, or 27 possible
codons.
20. Using the genetic code given in Figure 15.14, give the amino acids specified by the
bacterial mRNA sequences and indicate the amino and carboxyl ends of the
polypeptide produced.
Each of the mRNA sequences begins with the three nucleotides AUG. This indicates the
start point for translation and allows for a reading frame to be set. In bacteria, the
AUG initiation codon codes for N-formyl-methionine. Also for each of these mRNA
sequences, a stop codon is present either at the end of the sequence or within the
interior of the sequence.
The amino terminal refers to the end of the protein with a free amino group and will
be the first peptide in the chain. The carboxyl terminal refers to the end of the protein
with a free carboxyl group and is the last amino acid in the chain. For the following
peptide chains reading from left to right, the first amino acid is located at the amino
end, while the last amino acid is located at the carboxyl end.
(a) 5'–AUGUUUAAAUUUAAAUUUUGA–3'
Amino fMet–Phe–Lys–Phe–Lys–Phe Carboxyl
(b) 5'–AUGUAUAUAUAUAUAUGA--3'
Amino fMet–Tyr–Ile--Tyr--Ile Carboxyl
(c) 5'–AUGGAUGAAAGAUUUCUCGCUUGA–3'
Amino fMet–Asp–Glu–Arg–Phe–Leu–Ala Carboxyl
(d) 5'–AUGGGUUAGGGGACAUCAUUUUGA–3'
Amino fMet–Gly Carboxyl (The stop codon UAG occurs after the codon for
glycine.)
21. A nontemplate strand on DNA has the following base sequence. What amino acid
sequence would be encoded by this sequence?
5'–ATGATACTAAGGCCC–3'
To determine the amino acid sequence, we need to know the mRNA sequence and the
codons present. The nontemplate strand of the DNA has the same sequence as the
mRNA except that thymine containing nucleotides are substituted for the uracil
containing nucleotides. So the mRNA sequence would be as follows:
5'–AUGAUACUAAGGCCC–3'.
Assuming that the AUG indicates a start codon, then the amino acid sequence
would be starting from the amino end of the peptide and ending with the carboxyl end:
fMet–Met–Leu–Arg–Pro
194 Chapter Fifteen: The Genetic Code and Translation
22.
The following amino acid sequence is found in a tripeptide: Met–Trp–His.
Give all possible nucleotide sequences on the mRNA, on the template strand of DNA,
and on the nontemplate strand of DNA that could encode this tripeptide.
The potential mRNA nucleotide sequences encoding for the tripeptide Met–Trp–His can
be determined by using the codon table found in Figure 15.14. From the table, we can
see that the amino acid His has two potential codons, while the amino acids Met and
Trp each have only one potential codon. So, there are two different mRNA nucleotide
sequences that could encode for the tripeptide. Once the potential mRNA nucleotide
sequences have been determined, the template and nontemplate DNA strands can be
derived from these potential mRNA sequences.
(1) 5'–AUGUGGCAU–3'
DNA template:
DNA nontemplate:
(2) 5'–AUGUGGCAC–3'
DNA template:
DNA nontemplate:
3'–TACACCGTA–5'
5'–ATGUGGCAT–3'
3'–TACACCGTG–5'
5'–ATGTGGCAC–3'
23.
How many different mRNA sequences can code for a polypeptide chain with the amino
acid sequence Met–Leu–Arg? (Be sure to include the stop codon.)
From Figure 15.14, we can determine that leucine and arginine each have six different
potential codons. There are also three potential stop codons. As for methionine, only
one codon, AUG, is typically found as the initiation codon. (However, both UUG and
GUG have been shown to serve as start codons on occasion. For this problem, we will
ignore these rare cases.) So, the number of potential sequences is the product of the
number of different potential codons for this tripeptide, which gives us a total of (1 × 6
× 6 × 3) = 108 different mRNA sequences that can code for the tripeptide Met–Leu–
Arg.
24.
A series of tRNAs have the following anticodons. Consider the wobble rules given in
Table 15.2, and give all possible codons with which each tRNA can pair.
From the wobble rules outlined in Table 15.2, we can see that when “A” occurs at the
5' of the anticodon it can only pair with “U” in the 3' end of the codon. When “C” is
present at the 5' of the anticodon, it can only pair with “G” at the 3' of the codon.
However, both “U” and “G” when present at the 5' end of the anticodon can pair with
two different nucleotides at the 3' end of the codon (U with A or G; and G with U or C).
The rare base iosine (I) is also found at the 5' of the anticodon of tRNA on occasion.
Iosine can pair with “A”, “U,” or “C” at the 3' end of the codon.
(a) 5'–GGC–3'
Codons: 3'–CCG–5' or 3'–UCG–5'.
(b) 5'–AAG–3'
Codon: 3'–UUC–5'.
(c) 5'–IAA–3'
Codons: 3'–AUU–5' or 3'–UUU–5' or 3'–CUU–5'.
(d) 5'–UGG–3'
Codons: 3'–ACC–5' or 3'–GCC–5'.
Chapter Fifteen: The Genetic Code and Translation 195
(e) 5'–CAG–3'
Codon: 3'–GUC–5'.
25. An anticodon on a tRNA has the sequence 5'–GCA–3'.
(a) What amino acid is carried by this tRNA?
The anticodon 5'–GCA–3' would pair with the codon 5'–CGU–3'. Based on the
codon table in Figure 15.14, the amino acid encoded by this codon is cysteine. So,
this tRNA is most likely carrying cysteine.
(b) What would be effect if the G in the anticodon were mutated to a U?
The anticodon would now be 3'–ACU–5' and could pair to the codon 5'–UGA–3', a
stop codon. The result would be that amino acid cysteine would be placed where the
stop codon 5'–UGA–3' was located in the mRNA. Essentially, the stop codon would
be suppressed and translation could continue.
26. Which of the following amino acid changes could result from a mutation that changed a
single base? For each change that could result from the alteration of a single base,
determine which position of the codon (first, second or third nucleotide) in the mRNA
must be altered for the change to occur.
(a) Leu  Gln
Of the six codons that encode for Leu, only two could be mutated by the alteration
of a single base to produce the codons for Gln:
CUA (Leu)—Change the second position to A to produce CAA (Gln).
CUG (Leu)—Change the second position to A to produce CAG (Gln).
(b) Phe  Ser
Both Phe codons (UUU and UUC) could be mutated at the second position to
produce Ser codons:
UUU (Phe)—Change the second position to C to produce UCU (Ser).
UUC (Phe)—Change the second postion to C to produce UCC (Ser).
(c) Phe  Ile
Both Phe codons (UUU and UUC) could be mutated at the first position to produce
Ile codons:
UUU (Phe)—Change the first position to A to produce AUU (Ile).
UUC (Phe)—Change the first position to A to produce AUC (Ile).
(d) Pro  Ala
All four codons for Pro can be mutated at the first position to produce Ala codons:
CCU (Pro)—Change the first position to G to produce GCU (Ala).
CCC (Pro)—Change the first position to G to produce GCC (Ala).
CCA (Pro)—Change the first position to G to produce GCA (Ala).
CCG (Pro)—Change the first position to G to produce GCG (Ala).
(e) Asn  Lys
Both codons for Asn can be mutated at a single position to produce Lys codons:
AAU (Asn)—Change the third position to A to produce AAA (Lys).
AAU (Asn)—Change the third position to G to produce AAG (Lys).
AAC (Asn)—Change the third postion to A to produce AAA (Lys).
AAC (Asn)—Change the third position to G to produce AAG (Lys).
196 Chapter Fifteen: The Genetic Code and Translation
(f) Ile  Asn
Only two of the three Ile codons can be mutated at a single position to produce Asn
codons:
AUU (Ile)—Change the second position to A to produce AAU (Asn).
AUC (Ile)—Change the second position to A to produce AAC (Asn).
27.
A synthetic mRNA added to a cell-free protein-synthesizing system produces a peptide
with the following amino acid sequence: Met–Pro–Ile–Ser–Ala. What would be the
effect on translation if the following components were omitted from the cell-free
protein-synthesizing system? What, if any, type of protein would be produced? Explain
your reasoning.
(a) Initiation factor 1
The lack of IF1 would decrease the amount of protein synthesized. IF1 promotes the
disassociation of the large and small ribosomal subunits. Translation initiation
would occur, but at a slower rate since more of the small ribosomal subunits would
be bound to the large ribosomal subunits.
(b) Initiation factor 2
No translation would occur. IF2 is necessary for translation initiation. The lack of
IF2 would prevent fMet-tRNAfmet from being delivered to the small ribosomal
subunit, thus blocking translation.
(c) Elongation factor Tu
Although translation initiation or delivery of the Met to the ribosome-mRNA
complex would occur, no further amino acids would be delivered to the ribosome.
EF-TU is necessary for elongation where it binds GTP and the charged tRNA. This
three-part complex enters the “A” site of the ribosome. If EF-TU is not present,
then the charged tRNA will not enter the “A” site, thus stopping translation.
(d) Elongation factor G
EF-G is necessary for the translocation of the ribosome along the mRNA in a 5' to
3' direction. Once the formation of the peptide bond occurs between the Met and
Pro, the lack of EF-G would prevent the movement of the ribosome along the
mRNA, so no new codons would be read. However, the dipeptide Met-Pro would be
formed since its formation would not require EF-G.
(f) Release factors R1, R2, and R3
The release factors recognize the stop codons and promote cleavage of the peptide
from the tRNA at the “P” site. The absence of the release factors prevents
termination of translation at the stop codon, resulting in a larger peptide.
(g) ATP
ATP is required for the charging of the tRNAs with amino acids by the aminoacyltRNA synthetases. Without ATP, the charging would not take place, and no amino
acids will be available for protein synthesis. So no protein synthesis will occur.
(h) GTP
GTP is required for initiation, elongation, and termination of translation. If GTP is
absent, no protein synthesis will occur.
Chapter Fifteen: The Genetic Code and Translation 197
CHALLENGE QUESTIONS
28. In what ways are spliceosomes and ribosomes similar? In what ways are they different?
Can you suggest some possible reasons for their similarities?
Spliceosomes and ribosomes are both large complexes that are composed of several
different RNA and protein molecules. In essence, both the spliceosome and ribosome
are RNA-based enzymes or ribozymes. The RNA molecules in both structures are
necessary for catalysis. The 23S RNA molecule in the ribosome catalyzes the formation
of peptide bonds between amino acids. Other rRNAs are also important for protein
synthesis. In spliceosomes, the snRNA molecules catalyze the cutting and splicing of
pre-mRNA molecules to produce mature mRNA. The catalytic RNAs of the ribosome
and the spliceosome may have originated during the RNA world when RNA molecules
served to store information and to catalyze reactions that sustained life.
29. Several experiments were conducted to obtain information about how the eukaryotic
ribosome recognizes the AUG start codon. In one experiment, the gene that codes for
methionine initiator tRNA (tRNAiMet) was located and changed. The nucleotides that
specify the anticodon on tRNAiMet were mutated so that the anticodon in the tRNA was
5'–CCA–3' instead of 5'–CAU–3'. When this mutated gene was placed into a eukaryotic
cell, protein synthesis took place, but the proteins produced were abnormal. Some of
the proteins produced contained extra amino acids, and others contained fewer amino
acids.
(a) What do these results indicate about how the ribosome recognizes the starting point
for translation in eukaryotic cells? Explain your reasoning.
By mutating the anticodon to 5'–CCA–3' from 5'–CAU–3' on tRNAiMet, the initiator
tRNA will now recognize the codon 5'–UGG–3', which normally would code only
for Trp. If translation initiation by the ribosome in eukaroytes occurs by binding the
5' cap of the mRNA followed by scanning, then the first 5'–UGG–3' codon
recognized by the mutated tRNAiMet will be the start site for translation. If the first
5'–UGG–3' codon occurs prior to the normal 5'–AUG–3' codon, then a protein
containing extra amino acids could be produced. If the first 5'–UGG–3' codon
occurs after the normal 5'–AUG–3', then a shorter protein will be produced.
Finally, truncated proteins could also be produced by the first 5'–UGG–3' being out
of frame of the normal coding sequence. If this happens, then most likely a stop
codon will be encountered before the end of the normal coding sequence and will
terminate translation. The data suggest that translation initiation takes place by
scanning of the ribosome for the appropriate start sequence.
(b) If the same experiment had been conducted on bacterial cells, what results would
you expect?
Very little or no protein synthesis would be expected. Translation initiation in
bacteria requires the 16S RNA of the small ribosomal subunit to interact with the
Shine-Dalgarno sequence. This interaction serves to line up the ribosome over the
start codon. If the anticodon has been changed such that the start codon cannot be
recognized, then protein synthesis is not likely to take place.