Download Answers questions chapter 15

Chapter 15
Translation
a. What is the Shine-Dalgarno sequence? What binds to the sequence during translation? Describe some
of the ways in which the sequence can vary, addressing the consequences that any such variation has on
translation efficiency.
Suggested Answer:
The Shine-Dalgarno sequence, also called the ribosome binding site, is a sequence element (ideally, 5′GGAGG-3′) that is found just upstream (usually, 3 to 9 nucleotides away) of the start codon in many, but
not all, prokaryotic genes.
The sequence is recognized by the small ribosomal subunit (specifically, by a region of the 16S RNA that
contains the complementary sequence 5′-CCUCC-3′), allowing the subunit to bind to the mRNA near the
start codon with the proper positioning for the initiation of translation.
The Shine-Dalgarno sequence can vary considerably, in terms of both its sequence and as its distance
from the start codon. Sequences that are most similar to the ideal sequence—which is perfectly
complementary to the 16S RNA sequence—tend to have the most activity, while sequences that deviate
from the ideal are generally less active. Also, sequences that have the proper spacing relative to the start
codon (i.e., 3 to 9 bases away) are typically more active than those at other locations.
b. What is the role of the 5′ cap during eukaryotic translation? What happens to mRNA molecules that are
missing a cap?
Suggested Answer:
eIFs recognize the 5′ cap and then recruit the small subunit of the ribosome to the mRNA. Once bound to
the cap, the small subunit scans along the RNA until it encounters a start codon.
The cap also serves to protect mRNA from degradation in the cell, as RNAs lacking a cap are rapidly
destroyed by 5′ to 3′ exonucleases that are unable to cleave the 5′-5′ bond found in the cap.
c. Describe the structural and sequence elements that are common to all tRNA molecules, addressing the
function of each of the elements. What forces stabilize the tRNAs' structural features?
Suggested Answer:
tRNAs all share a secondary structure that resembles a cloverleaf, including a stem, three stem-loops,
and a variable loop. The stem of the cloverleaf, called the acceptor stem, is made up of the 5′ and 3′ ends
of the tRNA, and is the site of amino acid attachment to the tRNA. The three stem-loops include the ΨU
loop, which usually contains a pseudouridine base; the D loop, which contains dihydrouridine bases; and
the anticodon loop, which includes the three-nucleotide anticodon sequence that forms base pairs with
codons in the mRNA. The variable loop can be anywhere from 3 to 21 nucleotides in size. The tertiary
structure of these domains assumes an L shape, in which the acceptor stem and the stem of the ΨU loop
form one extended helix, and the anticodon stem and the stem of the D loop form a second helix that
extends away from the first helix at a 90-degree angle.
Several kinds of interactions help stabilize the tRNA structure, including base pairing (including nonWatson-Crick pairing) within the helical regions, stacking interactions between the bases of the base
paired regions, and other interactions between the bases and the sugar-phosphate backbone.
d. Outline the steps by which aminoacyl tRNA synthetases charge tRNAs. How can some organisms get
away with having fewer than 20 synthetases, yet still charge tRNAs with all 20 amino acids?
Suggested Answer:
Aminoacyl tRNA synthetases charge tRNAs in two steps. In the first step, called adenylation, the amino
acid reacts with ATP in a way that adds an AMP molecule (through its phosphate group) to the carbonyl
group of the amino acid. This reaction also releases pyrophosphate, which is then hydrolyzed to help
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drive the overall reaction forward. In the second step, called tRNA charging, the adenylated amino acid
reacts with tRNA, transferring the amino acid to the 3′ end of the tRNA (to either the 2′ or 3′ hydroxyl
group) and releasing AMP in the process.
A given tRNA synthetase can charge only a single kind of amino acid, which typically means that an
organism has 20 different synthetase enzymes. Some organisms, however, have fewer than 20 tRNA
synthetases, although they still manage to charge tRNAs with all 20 amino acids. They can do this by
initially charging the tRNAs with fewer than 20 amino acids (using the available tRNA synthetases) and
then using other enzymes to modify some of the charged amino acids to re-create all 20 amino acids. For
example, some bacteria lack a synthetase for charging tRNAs with the amino acid glutamine, and charge
Gln
tRNA in the cell with glutamic acid instead. A second enzyme then modifies the glutamic acid attached
Gln
Glu
to the tRNA molecules (but not the glutamic acid linked to tRNA ) to convert the glutamic acid to
glutamine.
e. Outline the steps of the ribosome cycle. At what stage do the ribosomal subunits bind to each other?
To mRNA? What causes them to dissociate when protein synthesis is complete?
Suggested Answer:
The ribosome cycle begins when the small subunit of the ribosome binds to an mRNA. The small subunit
then recruits a large subunit to form a complete ribosome around the mRNA. The ribosome then begins
protein synthesis, moving from the start codon in the 5′ to 3′ direction along the mRNA, adding one amino
acid to the growing polypeptide chain for each successive codon on the RNA. When it reaches a stop
codon, the ribosome releases the polypeptide chain, releases the mRNA, and dissociates to reproduce
the large and small subunits.
f. Describe the A, P, and E sites of the ribosome. Where are they located? What occupies each of the
sites during the following stages of translation:
• upon creation of the 70s initiation complex
• after an initial round of peptidyl transfer, prior to aminoacyl tRNA binding
• following aminoacyl tRNA binding, but before the peptidyl transfer reaction
• after peptidyl transfer, but before translocation is completed by EF-G
• in the presence of puromycin, prior to peptidyl transfer
• upon reaching a stop codon
• following release of the polypeptide chain, but before the appearance of RRF
Suggested Answer:
The A site of the ribosome binds to the aminoacylated tRNA (i.e., the charged tRNA), the P site binds the
peptidyl-tRNA (i.e., the tRNA to which the polypeptide chain is linked), and the E (for exit) site is occupied
by the deacylated (i.e., uncharged) tRNA from which the growing polypeptide chain has just been
transferred (see figure below). Each of the three sites is formed at the interface between the large and the
small ribosomal subunits.
The A, P, and E sites are occupied in the following ways at the indicated stages of translation:
fMet
• A, E sites empty, P site containing fMet-tRNAi
fMet
• A site empty, P site containing peptidyl-tRNA, E site containing deacylated tRNAi
• A site containing aminoacyl tRNA, P site containing peptidyl-tRNA, E site containing deacylated
tRNA
• A site of large subunit empty, A site of small subunit containing anticodon end of peptidyl-tRNA; P
site of large subunit containing protein-linked end of peptidyl-tRNA, P site of small subunit
containing anticodon end of deacylated-tRNA; E site of large subunit containing acceptor stem
end of deacylated tRNA, E site of small subunit containing anticodon end of deacylated tRNA;
• A site containing puromycin, P site containing peptidyl-tRNA, E site containing deacylated tRNA;
• A site containing release factor, P site containing peptidyl tRNA, E site containing deacylated
tRNA;
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•
A site empty, P, E sites containing deacylated tRNA.
g. Describe the different mechanisms used to ensure translation accuracy, beginning with the charging of
the tRNA and continuing up to the transfer of the amino acid to the growing polypeptide chain. How does
the degree of translation accuracy compare to that of DNA replication or of transcription? Does it make
sense that the rate should be lower for translation than it is for these other processes?
Suggested Answer:
The ribosome contains three proofreading activities that help it ensure the accuracy of amino acid
incorporation. All of these act indirectly to ensure the accuracy of the incorporated amino acid, acting
specifically at the level of codon-anticodon pairing. The first of these mechanisms involves interactions
between two adjacent adenine residues of the 16S RNA and the minor groove of the codon-anticodon
complex. If (and only if) the base pairing of the codon and anticodon is correct, these residues help
stabilize the complex, increasing the affinity of the tRNA for the ribosome. The second mechanism
involves the GTPase activity of EF-Tu. This activity, which is required for EF-Tu to release the bound
tRNA and leave the ribosome, is stimulated only when a correct codon-anticodon match is present. The
third mechanism functions once EF-Tu has left and the aminoacyl tRNA occupies the A site. Here, the
aminoacyl tRNA must rotate into the peptidyl transferase center of the ribosomal large subunit in order for
peptidyl transferase to occur. This rotation, however, frequently causes tRNAs with incorrectly base
paired anticodons to dissociate from the ribosome.
–3
–4
The accuracy of translation (with an error rate of 10 to 10 ) is somewhat lower than that of transcription
–4 to
–5
–10
(10
10 ), and much lower than that of DNA replication (as low as 10 ). It makes sense that
translation has the highest error rate because not only is translation the most complex of the three
processes, involving more potential sources of errors, but also because mistakes that occur during
protein synthesis only affect a single protein. Errors that occur during transcription, on the other hand, are
already more serious, because a single mRNA can be used to synthesize multiple polypeptides. Finally, it
makes sense that DNA replication has the highest accuracy because the permanent nature of the DNA in
the nucleus means that any error that arises during replication will be present in all of the cell's
descendants.
h. How do prokaryotic cells recognize the stop codon during translation? How does recognition of the
codon lead to the termination of protein synthesis and dissociation of the ribosomal subunits? What
happens if the stop codon is missing?
Suggested Answer:
The stop codon is recognized in prokaryotic cells by the class I release factors RF1 and RF2 (UAG by
RF1, UGA by RF2, and UAA by both RF1 and RF2). These factors are composed entirely of protein and
use a particular stretch of three amino acids to form a "peptide anticodon" that binds to and specifically
recognizes the stop codon.
When the ribosome encounters a stop codon, RF1 or RF2 binds to the codon using its three amino acid
peptide anticodon, and another region of the protein (containing the short amino acid sequence GGQ)
interacts with the peptidyl transferase center of the ribosome to effect the hydrolysis of the bond linking
the polypeptide to the tRNA occupying the P site. Once the polypeptide has been released, a class II
release factor called RF3 binds to the ribosome. RF3 is a GTP binding protein and binds to the ribosome
in its GDP-bound form. The interaction with the ribosome causes RF3 to exchange GDP for GTP with the
result that RF3 binds more strongly to the ribosome and displaces the class I release factor. With the
class I release factor gone, the ribosome stimulates the GTPase activity of RF3, causing it to hydrolyze its
bound GTP and then let go of the ribosome. Next, a protein called RRF binds to the empty A site of the
ribosome, where it recruits EF-G to help it displace the deacylated tRNAs from the P and E sites. Once
the tRNAs leave the ribosome, RRF, EF-G, and the mRNA are all released by the ribosome. Finally, the
initiation factor IF3 helps dissociate the two ribosomal subunits from each other, thereby completing the
translation cycle.
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If the stop codon is missing, the ribosome translates to the end of the mRNA, passing through the poly-A
tail (and adding multiple lysines to the C terminus of the protein) and stalling once it reaches the very end.
A protein called Ski7 then dissociates the stalled ribosome from the mRNA and recruits an exonuclease
enzyme that degrades the defective mRNA. The protein is then also destroyed, as proteins containing
multiple lysines at their C termini are targeted for rapid degradation in the cell.
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