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MCAT Transcription, RNA Processing and Splicing Practice Questions

By StudyStore · Published 2026-10-20 · 8 original questions · about 12 min

Eight free MCAT-style practice questions on RNA polymerases, promoters, termination, capping, polyadenylation and splicing, with worked examples and full explanations.

Quick concept primer

Transcription copies the template strand into RNA in the 5' to 3' direction, so the RNA is complementary and antiparallel to the template and identical (with U in place of T) to the coding strand. In bacteria one RNA polymerase makes all RNA and a sigma factor recognizes promoter elements near -35 and -10. In eukaryotes, RNA polymerase I makes most rRNA, RNA polymerase II makes mRNA and some small RNAs, and RNA polymerase III makes tRNA and 5S rRNA. General transcription factors, including TBP binding the TATA box, assemble at the promoter.

Eukaryotic pre-mRNA is capped with 7-methylguanosine at the 5' end, spliced by the spliceosome to remove introns (which begin with GU and end with AG), and given a poly-A tail at the 3' end. Alternative splicing joins exons in different combinations, so one gene can encode several protein isoforms.

Traps: the template strand is complementary to the mRNA, not identical to it; and Shine-Dalgarno sequences belong to bacteria, while caps and tails belong to eukaryotes.

8 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

A segment of the template strand reads 3'-TACGGATC-5'. What is the sequence of the mRNA transcribed from it?

  1. 5'-GAUCCGUA-3'
  2. 5'-TACGGATC-3'
  3. 5'-UACGGAUC-3'
  4. 5'-AUGCCUAG-3'
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Answer: D. 5'-AUGCCUAG-3'

RNA is synthesized 5' to 3' and is complementary and antiparallel to the template. Pairing each template base reading from its 3' end gives A, U, G, C, C, U, A, G, so the mRNA is 5'-AUGCCUAG-3'. Choice C copies the template with U for T, which is not complementary, and choice B is DNA, not RNA.

Question 2

A cell line is treated with a low concentration of alpha-amanitin, which selectively inhibits RNA polymerase II. Which RNA would be reduced the most?

  1. Ribosomal RNA from the nucleolus
  2. Mitochondrial rRNA
  3. Transfer RNA
  4. Messenger RNA
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Answer: D. Messenger RNA

RNA polymerase II transcribes protein-coding genes, so mRNA synthesis drops. Most rRNA comes from RNA polymerase I, tRNA and 5S rRNA from RNA polymerase III (inhibited only at much higher alpha-amanitin concentrations), and mitochondrial RNA from a separate mitochondrial polymerase.

Question 3

A point mutation changes the TATA box of a eukaryotic gene to a sequence not recognized by TATA-binding protein. What is the most likely result?

  1. Reduced initiation of transcription from that promoter
  2. Increased transcription, because the polymerase is no longer blocked
  3. A premature stop codon in the mRNA
  4. Retention of introns
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Answer: A. Reduced initiation of transcription from that promoter

TBP, part of TFIID, binds the TATA box and nucleates assembly of the preinitiation complex including RNA polymerase II. Weaker TBP binding lowers the rate of initiation, so less mRNA is made. The mutation is in the promoter, not the coding region, so the mRNA sequence and splicing patterns are not directly affected.

Question 4

Each of the following is a step in eukaryotic mRNA maturation EXCEPT

  1. addition of a poly-A tail to the 3' end
  2. removal of introns by the spliceosome
  3. addition of a Shine-Dalgarno sequence upstream of the start codon
  4. addition of a 7-methylguanosine cap to the 5' end
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Answer: C. addition of a Shine-Dalgarno sequence upstream of the start codon

The Shine-Dalgarno sequence is a ribosome binding site found in bacterial mRNAs, where it base-pairs with 16S rRNA. In eukaryotes the 5' cap serves a similar recruiting function for the small ribosomal subunit. Capping, polyadenylation and splicing all occur in the nucleus before export of the mature mRNA.

Question 5

A gene has five exons. Exons 1 and 5 are always included in the mature mRNA. Exons 2, 3 and 4 are each independently either included or skipped. How many different mRNA isoforms are possible from this gene, assuming every combination is allowed?

  1. 8
  2. 6
  3. 9
  4. 3
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Answer: A. 8

Each of the three optional exons has two possibilities (included or skipped), independent of the others, so there are 2 x 2 x 2 = 8 combinations. This illustrates how alternative splicing multiplies the number of protein isoforms per gene. Whether all combinations occur in a real cell depends on regulatory splicing factors.

Question 6

A mutation changes the invariant GU at the 5' end of an intron to AU. Which outcome is most likely?

  1. Normal splicing, since only the exon sequence matters
  2. Conversion of the intron into a coding exon with no other change
  3. Increased polyadenylation efficiency
  4. Failure to recognize the 5' splice site, causing exon skipping, intron retention or use of a cryptic site, and usually an abnormal protein
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Answer: D. Failure to recognize the 5' splice site, causing exon skipping, intron retention or use of a cryptic site, and usually an abnormal protein

The GU at the intron 5' end and the AG at its 3' end are recognized by snRNPs of the spliceosome. A mutation at either position impairs splicing at that junction, so the spliceosome either skips the adjacent exon, retains the intron or uses a nearby cryptic site. All these outcomes typically alter the reading frame or insert extra sequence.

Question 7

Which feature is characteristic of intrinsic (rho-independent) termination of bacterial transcription?

  1. A GC-rich hairpin in the RNA followed by a run of uracils
  2. Binding of the TATA-binding protein
  3. Recognition of a stop codon by release factor
  4. A poly-A signal followed by cleavage and polyadenylation
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Answer: A. A GC-rich hairpin in the RNA followed by a run of uracils

The hairpin forms in the nascent RNA and stalls RNA polymerase, and the weak A-U base pairs in the stretch of uracils let the transcript dissociate from the template. Choice D describes eukaryotic mRNA 3' processing, and choice C describes termination of translation, not transcription.

Question 8

In a typical mammalian cell, which type of RNA makes up the largest fraction of total cellular RNA by mass?

  1. Messenger RNA
  2. MicroRNA
  3. Ribosomal RNA
  4. Transfer RNA
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Answer: C. Ribosomal RNA

Ribosomes are abundant, and rRNA is about 80 percent of the total RNA in a typical cell. Transfer RNA is a distant second, while mRNA, though diverse and functionally central, is only a few percent by mass because individual mRNAs are relatively short-lived and far fewer copies are present than of rRNA.

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.