Transcription and RNA Processing: Follow the Molecule

MCAT content library / Biology & biochemistry / Transcription and RNA Processing: Follow the Molecule

Keep DNA, precursor RNA, and processed RNA separate while solving strand-direction and exon–intron problems.

About 15 minutes, including the worked examples and checks.

From a gene to a functional result · Lesson 2 of 5 · Continue in this path

Useful foundations

Nucleotides, base pairing, and DNA stability

On this page

Identify which molecule is being changed
Work through a short template in its stated direction
Distinguish retained sequence from translated sequence
Worked application: explain a longer processed product
Use alternative splicing without inventing a DNA change
Check your reasoning
Continue learning
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Identify which molecule is being changed

Transcription makes an RNA copy from a DNA template. RNA processing then changes which sequences and end structures are present in the RNA product. These steps do not ordinarily remove introns from the genomic DNA. Keeping the DNA, primary transcript, and mature RNA separate prevents a short RNA band from being mistaken for a DNA deletion. AAMC includes transcription, RNA types, mRNA processing, and the machinery of splicing in Content Category 1B.¹

RNA polymerase reads its DNA template 3′ to 5′ while extending RNA 5′ to 3′. The newly synthesized RNA is complementary to the template. Across the same transcribed segment, it matches the coding DNA strand when both are written 5′ to 3′, except that RNA uses U instead of T. “Template” and “coding” describe the strands relative to a particular transcript; the top line of a diagram is not automatically the coding strand.²

For a nuclear protein-coding gene, transcription factors help recruit RNA polymerase II to the promoter. Initiation chooses where RNA synthesis begins, elongation extends the RNA, and termination ends transcription. A transcription start site is not a translation start codon: an mRNA can contain a 5′ untranslated region before the ribosome’s start site.³⁴

Work through a short template in its stated direction

The following sequence is an original internal segment of a transcribed region, not a complete gene. Its template is supplied in the direction in which the polymerase reads it.

Molecule	Aligned sequence
DNA template	3′-TGC ATA GGC-5′
New RNA	5′-ACG UAU CCG-3′
Coding DNA	5′-ACG TAT CCG-3′

One transcribed segment, with each strand direction stated

Pair the first template T with RNA A, the next G with C, and the next C with G. Continue across the segment to obtain ACG UAU CCG. The coding strand is a useful check: replacing its T with U gives the same RNA. If the template had instead been printed 5′ to 3′, the polymerase would traverse the displayed sequence in the opposite direction; copying complementary letters left to right would not give a 5′-to-3′ RNA answer.

Distinguish retained sequence from translated sequence

A typical eukaryotic nuclear pre-mRNA receives a modified guanosine cap at its 5′ end. Its 3′ end is cleaved, and poly-A polymerase adds a tail of A residues. These structures support stability, export, and translation. The added tail is not simply a long run copied from thymidines in the DNA template. Many processing events begin while transcription is still underway; the steps need not wait for one complete unprocessed molecule.⁴⁵

Introns are intervening sequences removed during a particular splicing event. Exons contribute sequence to the resulting spliced RNA. An exon can include a 5′ or 3′ untranslated region, so “exon” and “protein-coding sequence” are not interchangeable. Likewise, an intron is transcribed into the precursor even though it is absent from that mature product.⁴

The spliceosome is an RNA–protein assembly. Small nuclear RNAs, or snRNAs, associate with proteins in small nuclear ribonucleoprotein particles, or snRNPs. Together with other components, they recognize splice signals and carry out intron removal and exon joining. RNA participates in recognition and catalysis; a ribozyme is an RNA with catalytic activity. Some introns can self-splice, but ordinary spliceosome-dependent processing should not be pictured as an isolated RNA cutting itself without its machinery.⁵

Not every transcript is an mRNA waiting to become a protein. rRNAs form structural and catalytic parts of ribosomes; tRNAs serve as amino-acid adaptors during translation. These RNAs are themselves functional products and undergo processing. Their importance does not depend on being translated into polypeptides.⁶

Worked application: explain a longer processed product

An original illustrative gene contains the regions below. E denotes an exon and I an intron. A size assay reports the gene-derived sequence from the start of E1 through the end of E3, excluding the cap and poly-A tail. All products share those endpoints. The experiment separately checks genomic sequence, precursor RNA, and the junctions present in processed RNA.

Region	Length (nt)	Control processed RNA
E1	120	Retained
I1	80	Removed
E2	90	Retained
I2	110	Removed
E3	150	Retained

Regions in the order transcribed, from 5′ to 3′ in the RNA

Original region map on one shared length scale. E = exon; I = intron. Caps and poly-A tails are excluded from every total.

The unspliced precursor contains 120 + 80 + 90 + 110 + 150 = 550 nt. Removing both introns leaves 120 + 90 + 150 = 360 nt. A variant affecting a splice-recognition site at I1 produces a 440-nt RNA while the genomic region and full precursor remain 550 nt. The excess over the control processed product is 440 − 360 = 80 nt, exactly I1’s length. Retention of I1 is therefore a specific candidate explanation.

Length alone does not identify the retained sequence. Another sequence change could produce the same total length. Here, junction sequencing also detects E1–I1 and I1–E2 continuity, together with an E2–E3 junction that lacks I2. Those results confirm the depicted structure: I1 remains and I2 has been removed. The inference comes from agreement between the length difference and sequence-specific junction evidence.

A splice-site change does not always cause this outcome. It can lead to intron retention, exon skipping, or use of another splice site. The observed product must decide among them. Nor does an RNA of the expected size establish a functional protein: export, RNA stability, reading frame, premature stop signals, and protein activity require their own evidence.

Use alternative splicing without inventing a DNA change

Different splice choices can produce different RNAs from the same genomic sequence. Some products encode different protein forms; others change untranslated regions or the amount of usable RNA. Intronic sequences can contribute splice-recognition and regulatory signals, so “removed from this product” does not mean “biologically irrelevant.” Alternative splicing expands possible products without requiring an additional copy of the gene.⁵⁶

To evaluate a new result, first name the molecule measured. Then mark the shared endpoints and add the lengths of the retained regions. Finally, ask whether the evidence identifies a sequence, a total length, or an abundance. A shorter RNA, less RNA, and less DNA are three different observations. Moving between them requires an additional measurement or an explicitly supplied mechanism.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

An internal DNA template segment is printed 5′-CAGTTC-3′. RNA polymerase traverses this displayed segment from right to left. Which RNA sequence is produced, written 5′ to 3′?

A. 5′-CAGUUC-3′

Review this choice. This replaces T with U in the template itself. That shortcut applies to a coding strand written in the RNA’s direction, not to the supplied template.

B. 5′-CUUGAC-3′

Review this choice. This reverses the displayed order and replaces T with U but does not complement the template bases.

C. 5′-GAACUG-3′

Correct. Read the template right to left, from its 3′ end: C T T G A C. The growing RNA is G A A C U G, written 5′ to 3′.

D. 5′-GUCAAG-3′

Review this choice. These are complementary letters copied left to right. In that alignment the RNA runs 3′ to 5′, so it must be reversed before reporting a 5′-to-3′ answer.

The principle to carry forward

Template recognition and polymerase direction are separate checks. Label the ends before writing the RNA.

Check 2

A different precursor has regions E1 = 100 nt, I1 = 60 nt, E2 = 75 nt, I2 = 90 nt, and E3 = 125 nt, in that order. Both compared products retain the same full E1 and E3 endpoints; cap and tail lengths are excluded. The control removes both introns and retains every exon. A variant product is 225 nt. Among the four specified possibilities below, which fits its length?

A. E2 is skipped and both introns are removed.

Correct. The control is 100 + 75 + 125 = 300 nt. Skipping E2 removes 75 nt, leaving 225 nt.

B. I1 is retained and all exons are retained.

Review this choice. Retention of I1 would give 300 + 60 = 360 nt, not 225 nt.

C. I2 is retained and all exons are retained.

Review this choice. Retention of I2 would give 300 + 90 = 390 nt.

D. Neither intron is removed and all exons are retained.

Review this choice. Keeping every region would give the full 450-nt precursor.

The principle to carry forward

Compare products with the same assay endpoints. Add or subtract the candidate region, then use sequence evidence to confirm its identity.

Check 3

Two cell types have identical genomic sequences for gene R and make the same full precursor RNA. Junction sequencing shows that cell type A joins E1–E2–E3, whereas cell type B joins E1 directly to E3. Both RNAs have the same transcription start and 3′ cleavage sites. Which conclusion is supported by these observations?

A. Different DNA replication products account for the two RNA junction patterns.

Review this choice. The genomic sequences are supplied as identical. Different RNA junctions do not require different replicated DNA products.

B. Different poly-A tail lengths account for the missing E2 sequence.

Review this choice. A change at the added 3′ tail cannot create an E1–E3 junction inside the gene-derived RNA.

C. Different splice choices produce the two RNAs from the same precursor sequence.

Correct. The shared precursor and different exon junctions identify alternative RNA processing. E2 is retained in A and skipped in B.

D. Different transcription start sites account for the missing E2 sequence.

Review this choice. The start sites are explicitly identical, and the difference is an internal junction rather than loss of the transcript’s beginning.

The principle to carry forward

A changed RNA connection can arise through processing while the DNA and the initial transcript sequence stay the same.