Download Answers questions chapter 14

Chapter 14
RNA splicing
a. Consider the reactants and the products of a spliceosome-mediated splicing reaction. What
sequences/sites are necessary for splicing to occur? What are the consensus sequences at these sites?
Are these consensus sequences found in the exon or the intron? Explain.
Suggested Answer:
The 5′ splice site, the 3′ splice site, the branch site, and the polypyrimidine tract are necessary for a
splicing reaction to occur. The consensus sequences for the various sites are shown in Figure 14-3. The
consensus sequences are found primarily in the intron because the sequence of the exon is devoted to
encoding for the specific gene.
b. Describe the reaction that occurs during splicing to form a lariat.
Suggested Answer:
A lariat is the shape of the released intron that was joined to the branch site during the first
transesterification reaction during splicing. The splicing reaction consists of two successive
transesterification reactions, in which the first one is an SN2 reaction, where the 2′-OH of the A at the
branch site acts as a nucleophile and attacks the phosphoryl group of the G in the 5′ splice site. The
consequence of this reaction is that the 5′ junction between the exon and intron is cleaved and the 5′ end
of the intron is joined to the A within the branch site. The second transesterification reaction cleaves the 3′
end of the intron from the adjacent exon and the intron is released in the form of a lariat.
c. Describe the function of each of the following spliceosome components within the splicing process:
U1
U2
U4
U5
U6
U2AF
BBP
Suggested Answer:
U1: U1 acts at the beginning of the splicing pathway to recognize the 5′ splice site, using its RNA
component to hybridize to the mRNA. Subsequently, once the tri-snRNP particle is formed, U1 leaves the
complex and is replaced by U6.
U2: Following the formation of the Early complex, U2 displaces the BBP at the branch site, base pairing in
a way that exposes the branch site A so that it is available to attack the 5′ splice site.
U4: Once the A complex is formed, U4 joins the spliceosome as part of the tri-SNP particle (which also
includes U5 and U6), forming the B complex and bringing all three splice sites together. Later, when the
assembly pathway is complete, U4 leaves the spliceosome, allowing U6 to interact with U2 to form the
active site.
U5: U5 joins the spliceosome as part of the tri-snRNP particle, bringing the three splice sites together and
forming the B complex. U5 also helps bring the two exons together to facilitate the second splicing
reaction.
U6: U6 also joins the spliceosome as part of the tri-snRNP particle. Once the B complex has been
formed, U1 leaves the complex and is replaced by U6 at the 5′ splice site. Subsequently, U4 also leaves,
allowing U6 to interact with U2 to form the active site.
U2AF: U2AF binds early during the splicing pathway as a dimer with one subunit binding to the Py tract
and both interacting with the 3′ splice site and helping BBP bind to the branch site. Later, U2AF also helps
U2 bind to the branch site, displacing BBP.
BBP: BBP binds to the branch site at an early stage in the splicing pathway, contributing to the Early
complex. Later, following the formation of the A complex, BBP is displaced from the branch site by U2
d. Compare the self-splicing reactions carried out by group I and group II introns. Which is most similar to
the reaction carried out by the spliceosome on pre-mRNAs? Are these self-splicing introns enzymes?
Why or why not?
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Suggested Answer:
Group II introns carry out self-splicing using the same chemistry as that used by the spliceosome during
mRNA splicing. Specifically, an A residue at the branch point uses its 2′-OH to attack the phosphoryl
group of a G at the 5′ splice site. This links the branch site A to the 5′ splice site G, cleaving the
phosphodiester linkage at the 5′ splice site and freeing the 3′ end of the (5′) exon. This free 3′ end then
attacks the phosphoryl group at the 3′ splice site, joining the two exons together and releasing the intron
in the form of a lariat.
Group I introns, in contrast, use a free G nucleotide (instead of a branch site A) to attack the 5′ splice site.
As with group II introns, this cleaves the phosphodiester bond at the 5′ splice site, freeing the 3′ end of the
exon to attack the 3′ splice site. In this way, the exons are linked together and the intron is released
(although here the intron is linear, not as a lariat as with group II introns).
Self-splicing introns are not enzymes because they mediate only a single round of splicing, in contrast to
true enzymes which are capable of repeatedly catalyzing a reaction. A group I intron can be readily
converted into an enzyme, however, by providing it with a free G and allowing it to react with molecules
having a sequence complementary to its internal guide sequence; in this way, the intron can act in trans
to cleave an unlimited number of heterologous substrates.
e. How many exons does the average human gene have? How large are typical human exons and
introns? What significance does the relative size of exons and introns have for the evolution of genes and
of gene function?
Suggested Answer:
The average human gene includes 8 or 9 exons, with the exons averaging about 150 nucleotides in
length and the introns about 3,500.
The relative size of introns and exons--and in particular the fact that introns are much larger than exons-has important implications for the evolution of genes and of gene function. Because many functional and
structural domains are confined to single exons, and because the greater length of introns means that
many more recombination events occur within introns than they do within exons, recombination tends to
shuffle entire exons to create new combinations of functional and structural domains. Throughout
evolution, some of the new combinations created in this way were undoubtedly useful, helping to explain
why many individual proteins include repeated domains, and why many unrelated genes share similar
domains.
f. How does the number and size of introns in the average gene make splice site selection such a
challenge for cells? Describe two different mechanisms that cells use to increase the accuracy of splice
site selection.
Suggested Answer:
Accurate splicing requires that the 5′ splice site pairs with the correct 3′ splice site. While this may be
relatively straightforward for genes containing a small number of modestly sized introns, it can present a
huge challenge for those that include large numbers of introns (for example, well over a hundred in some
cases), and/or introns of vast dimensions (for example, up to 500,000 nucleotides). This challenge is
compounded by the relatively weak sequence requirements for splice sites, as this increases the
probability that random sequences will be mistakenly used for splicing. The cell has two mechanisms for
getting around these difficulties and increasing the accuracy of splice site selection.
The first mechanism acts to reduce the possibility that exons are skipped by the splicing machinery. In
this case, splicing factors associate with RNA polymerase during transcription, accompanying the
polymerase as it progresses along the DNA template. When the polymerase transcribes a 5′ splice site,
the factors leave the polymerase and bind to the RNA. They then wait until a 3′ splice site is transcribed
and then interact with factors that bind there, helping to ensure that the 3′ site is used.
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The second mechanism is designed to ensure that true splice sites are used for splicing. Here, a set of
proteins, called SR proteins, bind to specific sequences that are located within exons, but close to the
intron boundary. Once bound, they help recruit the splicing machinery, thereby ensuring that splicing
occurs at sites close to exon-intron boundaries (where it should occur) rather than at cryptic sites located
far from any exons.
g. Describe the two types of RNA editing, outlining the different steps involved in each of them. Which of
these involves the most significant changes in the mRNA sequence?
Suggested Answer:
The first type of RNA editing is called site-specific deamination. In this process, which only affects certain
genes in specific cell types, an enzyme called a deaminase removes amino groups from particular
nucleotides within the mRNA. Deamination can alter the identity of the affected nucleotides, converting,
for example, a C to a U, or an A to an inosine. This can have important consequences for the sequence
of the encoded protein, causing amino acid substitutions in the protein or replacing amino acid encoding
codons with stop codons.
The second type of RNA editing, which affects the mRNA sequence more dramatically than deamination,
occurs in the mitochondria of trypanosomes. This type of editing involves specialized RNA molecules,
called guide RNAs, which are complementary to portions of the mRNA but which contain additional As in
their sequence. During editing, the guide RNAs hybridize to their target mRNAs, and an endonuclease
cuts the mRNA at a position opposite one of the additional As in the guide sequence. A U is then inserted
into the mRNA opposite the additional A. Editing can add a large number of Us to an mRNA, dramatically
changing the sequence of the encoded protein either through the addition of multiple amino acids or by
causing a frameshift mutation in the mRNA.
h. What signals an mRNA for transport?
Suggested Answer:
Mature mRNAs are processed by the cell and become associated with specific proteins that signal the
mRNA for transport. For example, the splicing regulators SR proteins and proteins that interact only at
exon-exon junctions after splicing become associated with mRNAs during processing and help to trigger
the mature mRNA for transport.
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