Transcription and RNA Processing: Study Guide (Biology 6.3)

From a gene to a working molecule

A gene is a stretch of DNA that carries the instructions for a product, usually a protein. But DNA stays inside the nucleus of a eukaryotic cell, and proteins are built on ribosomes in the cytoplasm. The cell solves this by making a copy of the gene in RNA and sending the copy out to the ribosomes. The DNA itself stays safe.

Information therefore flows in two steps:

1. Transcription: DNA → RNA. A section of DNA is copied into a matching RNA molecule. This happens in the nucleus in eukaryotes.
2. Translation: RNA → protein. The RNA message is read on a ribosome to join amino acids in the right order (Topic 6.4).

This one-way flow, DNA → RNA → protein, is often called the central dogma of molecular biology. (Retroviruses are an exception you will meet in Topic 6.4.)

Three kinds of RNA, three jobs

RNA is a single-stranded nucleic acid. Its nucleotides contain the sugar ribose and the bases adenine (A), uracil (U), guanine (G) and cytosine (C). Because it is single-stranded, an RNA molecule can fold back on itself. Short stretches of complementary bases pair up inside the same molecule, giving each RNA a particular shape. So two things decide what an RNA does: its base sequence and the shape that sequence folds into.

RNA	Job	How sequence and structure fit the job
Messenger RNA (mRNA)	Carries a copy of a gene’s information from DNA in the nucleus to a ribosome in the cytoplasm	Its base sequence is the message: read three bases at a time, it specifies the order of amino acids
Transfer RNA (tRNA)	Brings a specific amino acid to the ribosome and matches it to the mRNA	Folds into a compact shape by internal base pairing. One end carries an amino acid; a loop at the other end holds the anticodon, three bases that pair with a matching codon on the mRNA
Ribosomal RNA (rRNA)	A main building block of ribosomes, together with proteins	Folds into a stable framework that holds mRNA and tRNA in position during translation

There are many different tRNAs. Each kind carries one particular amino acid and has its own anticodon. During translation, tRNAs are recruited to the ribosome one at a time, so the amino acids join in the order set by the mRNA. That is how a gene’s base sequence becomes a protein’s primary structure.

Transcription: copying one strand

Only one of the two DNA strands is copied for any one gene.

• The template strand is the strand that RNA polymerase reads. The new RNA is complementary to it.
• The coding strand (also called the non-template strand) is the other strand. It is not read, but its sequence is the same as the RNA’s, except that DNA has T where RNA has U.

The enzyme that does the work is RNA polymerase. It binds to the DNA at the start of a gene, unwinds a short stretch so the bases are exposed, and builds the RNA one nucleotide at a time. Base-pairing rules decide which nucleotide is added each time:

Base on the template strand	Base added to the RNA
A	U (RNA has no thymine)
T	A
G	C
C	G

Direction matters

Nucleic acid strands have two different ends, 5′ and 3′ (Topic 1.6). RNA polymerase can only add a new nucleotide to the 3′ end of the growing RNA. So:

• the RNA is built in the 5′ → 3′ direction;
• the template strand is read in the 3′ → 5′ direction, because the two strands pair in opposite (antiparallel) directions;
• the 5′ end of the RNA is made first.

Some other features to note:

• RNA polymerase does not need a primer, unlike DNA polymerase in replication (Topic 6.2). It can start a new chain from nothing.
• Only a gene-sized section is copied, not the whole chromosome. Transcription starts at the beginning of a gene and stops at its end. (How the cell controls which genes are transcribed is Topic 6.5.)
• For different genes, either DNA strand can be the template. “Template” describes a strand’s role for one gene, not a fixed property of the strand.
• Many RNA polymerases can copy the same gene one after another, so a cell can make many RNA copies of one gene.

RNA processing in eukaryotes

In a eukaryotic cell, the first RNA copy of a protein-coding gene is called pre-mRNA (or the primary transcript). Before it leaves the nucleus, enzymes modify it in three ways.

1. A 5′ cap is added. A modified guanine nucleotide (the course calls it the GTP cap) is attached to the 5′ end. The cap helps the ribosome recognise the mRNA and bind to it so translation can start.
2. A poly-A tail is added. A long run of adenine (A) nucleotides is added to the 3′ end. The tail makes the mRNA more stable: it protects the message from enzymes that break RNA down, so the mRNA lasts longer and can be translated more times.
3. Introns are removed and exons are joined (splicing). A gene contains exons, the sections kept in the mature mRNA, separated by introns, the sections cut out. A complex of proteins and small RNAs (background: the spliceosome) cuts out each intron and joins the exons on either side.

A memory aid: exons are expressed; introns stay in the nucleus and are broken down.

The result is a shorter mature mRNA that is exported to the cytoplasm. Prokaryotic mRNA does not go through this processing. Bacterial genes rarely contain introns, and bacterial mRNA is used straight away.

Alternative splicing: one gene, several products

Splicing does not always join every exon. In alternative splicing, a cell keeps some exons and leaves out others. Different cell types, or the same cell at different times, can join different combinations of exons from the same pre-mRNA. Each combination is a different mature mRNA, so it can be translated into a different protein.

This helps explain a puzzle. Humans have roughly 20,000 protein-coding genes, yet their cells make many more different proteins than that. Alternative splicing is one of the main reasons.

Worked example 1: writing a transcript from the template strand

Question. The template strand for part of a gene is written below in the 5′ → 3′ direction:

5′-CTAGGCATTCGACAT-3′

(a) Write the mRNA made from this region, labelling its 5′ and 3′ ends. (b) Write the coding strand, labelling its ends.

(a) Step by step.

1. RNA polymerase reads the template 3′ → 5′. So first rewrite the template in that direction (reverse it, keeping each base): 3′-TACAGCTTACGGATC-5′.
2. Pair each template base with its RNA partner (A→U, T→A, G→C, C→G), reading left to right.
3. The RNA is antiparallel to the template, so the end opposite the template’s 3′ end is the RNA’s 5′ end:

mRNA: 5′-AUGUCGAAUGCCUAG-3′

(b) The coding strand is complementary to the template, so it has the same sequence as the mRNA with T for U:

coding strand: 5′-ATGTCGAATGCCTAG-3′

Check. Three quick tests: (i) the mRNA contains no T; (ii) both the mRNA and the coding strand are 15 nucleotides, the same as the template; (iii) the mRNA and the coding strand match letter for letter except U/T, and both read 5′ → 3′ from left to right.

The common error. If you pair the bases without reversing the template, you get 5′-GAUCCGUAAGCUGUA-3′. The bases pair correctly one by one, but the ends are labelled the wrong way round, so it is a different message. Always check that the template and the RNA are antiparallel.

Worked example 2: splicing and alternative splicing in numbers

Question. A fictional eukaryotic gene is transcribed into a pre-mRNA with four exons and three introns, in this order (lengths in nucleotides, nt):

Section	E1	I1	E2	I2	E3	I3	E4
Length / nt	120	850	96	1400	75	620	210

Ignore the cap and the poly-A tail.

(a) Calculate the length of the pre-mRNA. (b) In liver cells all four exons are kept (variant A). Calculate the length of the mature mRNA, and the percentage of the pre-mRNA that is removed. (c) In kidney cells exon 3 is skipped (variant B). Calculate the length of this mature mRNA, and explain how the two cell types can make different proteins from the same gene.

(a) Add all seven sections: exons 120 + 96 + 75 + 210 = 501 nt; introns 850 + 1400 + 620 = 2870 nt. Pre-mRNA = 501 + 2870 = 3371 nt.

(b) Variant A keeps all exons: 501 nt. Removed: 2870 ÷ 3371 × 100 = 85.1%. Most of this pre-mRNA is intron.

(c) Variant B = 501 − 75 = 426 nt.

Both cell types have exactly the same DNA and transcribe the same pre-mRNA. The difference arises during splicing: kidney cells leave exon 3 out. The two mature mRNAs carry different base sequences, so they are translated into proteins with different amino acid sequences. Variant B’s protein lacks the part encoded by exon 3. If that part forms, for example, a binding site, the two proteins could behave differently.

Interpretation. Exon 3 is 75 nt long, which is a multiple of 3 (75 ÷ 3 = 25). In Topic 6.4 you will see why that matters: mRNA is read three bases at a time, so skipping a multiple of 3 removes whole codons without scrambling the exons that follow.

Common misconceptions

• “Both DNA strands are transcribed to make one mRNA.” Only the template strand is read for a given gene. The RNA matches the coding strand (with U for T).
• “The mRNA is identical to the template strand.” It is complementary to the template and antiparallel to it. It is the coding strand it resembles.
• “RNA polymerase builds RNA 3′ → 5′ because it reads the template 3′ → 5′.” It reads 3′ → 5′ but builds 5′ → 3′, adding each nucleotide to the 3′ end of the growing chain.
• “Transcription copies the whole chromosome, like replication.” It copies one gene, or a small group of genes, at a time, and different genes are transcribed at different times.
• “RNA polymerase needs a primer, like DNA polymerase.” It does not; it starts new RNA chains by itself.
• “Introns are removed from the DNA.” Introns stay in the DNA of the gene. They are removed from the RNA copy.
• “Exons are always joined in the same way.” Alternative splicing joins different sets of exons, so one gene can give several mature mRNAs.
• “The poly-A tail helps the ribosome find the mRNA.” In this course, the 5′ cap is linked to ribosome recognition; the poly-A tail is linked to stability.
• “tRNA and rRNA are just leftovers; only mRNA matters.” All three are made by transcription and all three are needed to make a protein.

Where this leads

Transcription produces the messages; the next step is reading them. In Topic 6.4, Translation, you will see how ribosomes, rRNA and tRNA turn the base sequence of a mature mRNA into a polypeptide, codon by codon. If base pairing or the 5′/3′ direction feels shaky, revisit Topic 6.2, DNA Replication first. Test yourself with the practice questions, then use the revision notes and the checklist to consolidate.