Genetics and Molecular Biology · Unit 4: From gene to protein, and its control
RNA Processing: Caps, Tails and Splicing
In the eukaryotic nucleus, the RNA that polymerase II makes is not yet a message. This lesson follows the three changes that make it one: a cap on the 5' end, a cut and a poly(A) tail at the 3' end, and splicing, which removes introns and joins exons. It tells how split genes were found in adenovirus 2 in 1977, and shows how to predict, from exon lengths alone, what a splice-site mutation does to an mRNA and its reading frame.
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The RNA that polymerase II copies from a eukaryotic gene is not the RNA that a ribosome reads.
Take an imagined human gene with four exons and three introns that runs to 5,137 nucleotides from the start of transcription to the end of its last exon. The finished mRNA carries 737 of them, plus a cap and a tail that the gene never encoded. This lesson is about the gap: what is added, what is cut out, how, and what goes wrong when one base at a cut site changes.
The idea: a primary transcript is not yet a message
What RNA polymerase II makes in transcription is the primary transcript, or pre-mRNA. Three changes turn it into mRNA.
- A cap is added to the 5’ end.
- The 3’ end is cut and given a poly(A) tail.
- Splicing removes the introns, the stretches of the gene that are not kept, and joins the exons, the stretches that are.
Think of an editor cutting a film while it is still being shot: none of the three steps waits for transcription to end. Capping happens once about 25 to 30 nucleotides of RNA have emerged from the polymerase, and splicing of early introns often begins while later parts of the gene are still being copied.
Bacterial mRNAs get no cap and carry no spliceosomal introns (a few bacterial genes hold self-splicing introns), and ribosomes translate them while they are still being made. The nuclear envelope gives eukaryotes time to process the message before it meets a ribosome.
Caps and tails: marking both ends
The cap is a guanosine methylated at position 7 (7-methylguanosine, m7G) joined to the first transcribed nucleotide 5’ to 5’, through a bridge of three phosphates, where a normal backbone links 3’ to 5’. Three enzymes build it in order:
- An RNA triphosphatase removes the outermost (gamma) phosphate from the RNA’s 5’ end.
- A guanylyltransferase transfers GMP from GTP onto that end, through an intermediate in which GMP is bound to a lysine of the enzyme.
- A methyltransferase adds a methyl group to N7 of the guanine, taking it from S-adenosylmethionine (see nucleotide metabolism for GTP and its relatives). This is cap 0. In many organisms a further methyl on the 2’-OH of the first transcribed nucleotide makes cap 1.
The cap blocks 5’-to-3’ exonucleases. In the nucleus a cap-binding complex holds it and helps the mRNA’s export. In the cytoplasm the translation factor eIF4E grips it and recruits the machinery that starts translation.
The tail starts with a cut. Near the end of the gene the RNA carries the signal AAUAAA, read in mammals by a complex called CPSF; cross-linking with purified human CPSF (Chan and colleagues, 2014) showed its subunits CPSF30 and Wdr33 contacting AAUAAA directly. Another complex, CstF, binds a GU-rich stretch downstream. The RNA is cut between them, about 10 to 30 nucleotides past AAUAAA, by the nuclease CPSF73. Poly(A) polymerase then adds adenosines to the new 3’ end without copying any DNA. The tail is not in the gene.
Newly made tails in mammalian cells are usually given as about 200 to 250 adenosines; yeast tails are shorter. Tails also shrink with time: in L4-stage C. elegans larvae, sequencing of individual tails found a median of 57. Ask of any tail length which organism, and whether new RNA or the whole population was measured.
With poly(A)-binding proteins, the tail helps protect the mRNA and promote translation. It is not universal. The replication-dependent histone mRNAs of animals are cut but not normally polyadenylated; their 3’ ends form a stem-loop instead.
The 1977 surprise: genes in pieces
Until 1977 a gene was expected to be a continuous stretch of DNA, as in E. coli. The evidence against came from adenovirus 2, a DNA virus that grows in human cells, and from two groups.
At MIT, Susan Berget, Claire Moore and Phillip Sharp purified hexon mRNA, the message for the virus’s main coat protein, and paired it with mapped pieces of viral DNA. At Cold Spring Harbor, Louise Chow, Richard Gelinas, Thomas Broker and Richard Roberts paired viral DNA with RNA from the polysomes of infected cells. Both viewed RNA-DNA hybrids in the electron microscope. Chow and colleagues made R loops, in which RNA displaces one DNA strand and pairs with the other; Berget and colleagues used single-stranded DNA fragments.
If a gene were continuous, a message would pair with one unbroken stretch of DNA. The body of each message did. Its 5’ end did not. Chow and colleagues mapped a leader of 150 to 200 nucleotides, shared by many late mRNAs and copied from three separate regions of DNA, now called the tripartite leader. Berget and colleagues proposed that four separate segments may be joined to make hexon mRNA, which implied that RNA copied from the DNA between them is removed.
Split genes were soon found in animal cell genes, including globin genes. The 1993 Nobel Prize in Physiology or Medicine went to Roberts and Sharp for their discoveries of split genes. Introns are not junk: some carry sequences that control splicing or transcription.
The spliceosome’s chemistry
Most human introns of the major (U2-type) class follow the GU-AG rule: GU at the 5’ end, AG at the 3’ end. A third site lies inside the intron, usually a few dozen nucleotides upstream of the 3’ end: the branch point, which contains an adenosine. Between it and the AG lies a run of pyrimidines.
Splicing is two transesterifications, each swapping one phosphodiester bond for another. No bond is lost overall, so the chemistry itself needs no ATP; the spliceosome spends ATP on assembling and rearranging itself.
- The 2’-OH of the branch-point adenosine attacks the phosphate at the 5’ splice site. The upstream exon is freed, and the intron’s 5’ G is now joined to the branch A by a 2’-5’ bond, forming a loop with a tail: the lariat.
- The freed 3’-OH of the upstream exon attacks the phosphate at the 3’ splice site. The exons are joined and the lariat intron is released.
The spliceosome is built anew on each intron from five small nuclear ribonucleoproteins (snRNPs: U1, U2, U4, U5 and U6, each a small RNA with proteins) and many other proteins. U1 snRNA finds the 5’ splice site by base pairing. U2 pairs with the branch-point sequence, leaving the branch A bulged out. The U4/U6.U5 tri-snRNP then joins, U1 and U4 leave, and the helicase Brr2 unwinds U4 from U6 so that U6 can pair with U2 and the 5’ splice site. Cryo-EM structures of yeast and human spliceosomes show U6 snRNA positioning the catalytic metal ions, so the active site is largely RNA, held and rearranged by proteins.
That fits a 1982 result: Thomas Cech’s laboratory showed that the intron of Tetrahymena ribosomal RNA removes itself with no protein. This group I intron uses an external guanosine, not a branch A, as the first attacking group. Self-splicing group II introns, found in organelles and bacteria, form a lariat as the spliceosome does.
A second, minor spliceosome, built with U11, U12, U4atac and U6atac, removes a small class of introns, well under 1 percent of those in human genes.
One gene, several proteins, and what a splice mutation does
Because splicing chooses which pieces to join, one gene can make several mRNAs. Wang and colleagues (2008) classed the patterns they found in human transcriptomes into skipped exons, alternative 5’ or 3’ splice sites, mutually exclusive exons and retained introns, and estimated that 92 to 94 percent of human genes undergo alternative splicing. Regulation of these choices belongs to eukaryotic gene regulation.
The same logic predicts mutations. A change that weakens a splice site most often causes the exon to be skipped. Less often a cryptic splice site, a sequence not normally used, takes over, or the intron is kept. A cloned beta-thalassemia allele with a change at the GT of intron 2, expressed in HeLa cells, gave a main RNA that kept the first 47 nucleotides of intron 2 and a minor one that joined exon 1 to exon 3. Since 47 = 3 x 15 + 2, the first shifts the frame.
The worked problem uses one rule: a change in coding length keeps the downstream frame only if it is a multiple of three.
frame kept if (change in coding length) / 3 leaves remainder 0
Worked problem. An imagined human gene has four exons. Exon 1 is 150 nucleotides: a 60-nucleotide 5’ UTR, then 90 coding nucleotides that begin with the AUG. Exon 2 is 87, exon 3 is 100, and exon 4 is 400 (it holds the stop codon and the 3’ UTR). Introns 1, 2 and 3 are 900, 1,500 and 2,000. Ignore the cap and tail.
- Mature mRNA. 150 + 87 + 100 + 400 = 737 nucleotides. (Adding the introns, 900 + 1,500 + 2,000 = 4,400, gives the 5,137 of the pre-mRNA.)
- A 5’ splice-site mutation of intron 2 causes exon 2 to be skipped. In vertebrates an exon is often recognised as a unit by the sites at both its ends, so losing one site can drop the whole exon. Length: 737 - 87 = 650. Since 87 = 3 x 29, the frame is kept and the protein loses 29 amino acids.
- Exon 3 is skipped instead. Length: 737 - 100 = 637. Since 100 = 3 x 33 + 1, the frame shifts by one base from the exon 2 to exon 4 junction on, and a premature stop codon in exon 4 is likely, though not certain without the sequence. In mammalian cells, a premature stop in the last exon usually escapes nonsense-mediated decay that depends on exon junctions (covered in the genetic code and translation), so a truncated protein could be made.
- Intron 1 is retained. Length: 737 + 900 = 1,637. Such RNAs are often kept in the nucleus or degraded.
Rule to keep: the downstream frame is kept only if the change in coding length is a multiple of three. This is a sketch of the logic for an imagined gene, not a prediction for a real one; the sequence decides whether a stop codon appears.
Design problem
A patient variant changes intron 3 position +5 (the G of GU is +1) from G to A, in a gene expressed in muscle. Design a minigene experiment that tests whether the variant changes splicing, predict the band sizes for each outcome, and say what the result can and cannot tell you.
One way to do it
Clone exon 3, intron 3 and exon 4 into a splicing reporter plasmid between two vector exons (see hacking the cell), in two versions differing only at +5. Transfect each into a cell line, ideally muscle-derived, and run RT-PCR with primers in the vector exons.
With invented lengths (100 nucleotides of vector exon between the primers, exon 3 of 120, exon 4 of 150, intron 3 of 800): normal splicing gives 100 + 120 + 150 = 370 bp, skipping exon 3 gives 250, and retaining intron 3 gives 1,170. A nearby cryptic 5’ site gives a band slightly off 370 that only sequencing can resolve.
Controls: empty vector (100 bp); wild-type minigene; a known splice-null change, such as +1 G to A of the same intron; no reverse transcriptase, to rule out plasmid DNA as template. Sequence every band.
Limits: the minigene lacks the rest of the gene, and the cell line may lack splicing factors of muscle, so the result is evidence, not proof. Confirm in patient RNA if it is available. The assay describes splicing; it is not a clinical reading of the variant.
How we measure it
R-loop and RNA-DNA hybrid mapping by electron microscopy
Pair a purified RNA with its DNA under conditions that favour RNA-DNA hybrids, then view the molecules in the electron microscope. Paired stretches and the loops of unpaired DNA between them show which parts of the DNA are present in the RNA and which are left out.
Minigene splicing assay
Clone an exon with its flanking intron sequence, wild type or variant, into a reporter plasmid, put it into cultured cells, and read the spliced RNA by RT-PCR with primers in the flanking exons. The band sizes, confirmed by sequencing, tell you whether the exon was included, skipped, or joined at a new site.
RNA sequencing with splice-junction reads
Sequence short fragments of cDNA made from mRNA and map them to the genome. Reads that span two exons record which exons were joined, so the share of messages that include a given exon can be estimated across tissues.
Finish line
Test yourself
8 questions
Practice and MCAT-style, with explained answers.
Next · Lesson 18
The Genetic Code and Translation
Sixty-four codons, twenty amino acids, three stops: how was the dictionary read, and who reads it in the cell?
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