RNA processing | Anatomy2Medicine
RNA processing

RNA processing


  • RNA processing (eukaryotes)


      • Initial transcript is called heterogeneous nuclear RNA (hnRNA).
      • hnRNA is then modified and becomes mRNA.


  • The following processes occur in the nucleus following transcription:


  • Capping of 5′ end (addition of 7-methylguanosine cap)
  • Polyadenylation of 3′ end (≈ 200 A’s)
  • Splicing out of introns
    • Capped, tailed, and spliced transcript is called mRNA.
    • mRNA is transported out of the nucleus into the cytosol, where it is translated.


  • mRNA quality control occurs at cytoplasmic P-bodies, which contain exonucleases,


decapping enzymes, and microRNAs;

  • mRNAs may be stored in cytoplasmic P-bodies for future translation.
  • Poly-A polymerase does not require a template.
  • AAUAAA = polyadenylation signal.

Splicing of pre-mRNA

    • Primary transcript combines with small nuclear ribonucleoproteins (snRNPs) and other proteins to form spliceosome.
    • Lariat-shaped (looped) intermediate is generated.
    • Lariat is released to precisely remove intron and join 2 exons.
    • Antibodies to spliceosomal snRNPs (anti- Smith antibodies) are highly specific for SLE.


  • Anti-U1 RNP antibodies are highly associated with mixed connective tissue disease.



Exam MCQ

A careful analysis of cellular components discovers short-lived RNA species in which an adenine nucleotide is found with three phosphodiester bonds (linked to the 2′, 3′, and 5′ carbons). This transient structure is formed during which of the following processes?

(A) mRNA cap formation

(B) mRNA polyadenylation

(C) Splicing of hnRNA

(D) Transcription of microRNAs

(E) Transcription of rRNA

5 The answer is C:

Splicing of hnRNA. As seen on page 34, an adenine nucleotide in the middle of the intron is a required component for splicing to occur, and the sugar residue attached to this adenine is involved in three phosphodiester linkages; the normal 3′ and 5′ and also 2′ to the splice site. The resulting structure resembles a lariat. Such a structure does not form during capping, polyadenylation, or the normal transcription of genes. It is unique to the splicing mechanism.


Exam MCQ

A mutation to this sequence in eukaryotic mRNA will affect the process by which the 3 end poly A tail is added to the mRNA.

  1. CAAT
  2. CCA

Ans D


  • For addition of poly A tail the mRNA is cleaved downstream of a consensus sequence called the polyadenylation signal sequence ie AAUAAA found near the 3 end of RNA and the poly A tail is added to the new 3 end, so mutation in this site will lead affect the poly A tail addition.


    • CAAT, GGGGCG AND TATAAA are sequences found in promoter for RNA polymerase


  • CCA is added to the 3 end of tRNA by nucleotidyl transferase


Exam MCQ

Pick out the factor required for splicing to convert heterogenous nuclear RNA (hnRNA to messenger RNA (mRN A) in eukaryotes:

  1. Small nuclear ribonucleo protein (SnRNAs)
  2. 5′ Capping
  3. Poly A tail
  4. None of the above

ANS: A.   Small ribonuclear proteins (Sn RNPs) remove the introns. Introns are intervening nucleotide sequences in mRNA, not code for proteins. 5′ end of mRNA is capped with 7- methyl guanylate by a 5′ to 5′ triphosphate linkage. Poly A tails is 200-300 adenylate residues at 3′ end. Poly A is added after transcription to stabilize DNA. When mRNA enters into the cytoplasm, poly A tail is slowly shortened, when it is completely removed, the mRNA is rapidly degraded. Sn RNPs are formed by the association of small nuclear RNA (Sn RNA) with proteins. Sn RNP association with hn RNA at the exon – intron junction is known as spliceosome. Spliceosomes contain five Sn RNAs U, U2, U4 U5 andU6.

Exam MCQ

  1. Which is the most common post transcriptional modification occurring in both mRNA and tRNA in eukaryotes:
  2. 5′ Capping with 7-methyl guanosine
  3. Addition of CCA nucleotides to 3′ terminal end
  4. Removal of introns by splicing
  5. Removal of polyA tail before entering into the cytoplasam

ANS: C.  

The mechanism of removal of introns by splicing is different in messenger RNA (mRNA) and transfer RNA (tRNA), but in both the RNAs introns are removed by splicing.

5′ end of the mRNA is capped with 7-methyl guanosine but in tRNA at 5’end no modifications are seen.

3′ end of tRNA, added with CCA nudeotide is sequenced by the enzyme tRNA nucleotide transferase. Post transcriptional modification, wherein mRNA possesses poly- A tail at the 3′ end. When mRNA enters into the cytoplasam poly-A chain gets reduced.


Exam MCQ


Failure of an endonuclease to recognize the sequence AAUAAA in the 3N end of heterogeneous nuclear RNA will cause a defect in which of the following processes involving mRNA?


  1. A) Capping
  2. B) Hybridization
  3. C) Polyadenylation
  4. D) Splicing
  5. E) Transport

Ans C

Exam MCQ

What is added to the 3′-end of many eukaryotic mRNAs after transcription? A. introns

  1. a poly A tail
  2. a cap structure, consisting of a modified G nucleotide
  3. the trinucleotide 5′-CCA
  4. exons

Ans B. a poly A tail
A tail of A-nucleotides, generally 100-200 long, is added to the 3′-end of most eukaryotic pre- mRNAs. The poly A tail, which is not coded in the DNA, is also retained in the mRNA exported to the cytoplasm.

Exam MCQ

Exam MCQ

Which of the following is NOT a feature of eukaryotic gene expression?

  1. polycistronic mRNAs are very rare
  2. many genes are interrupted by noncoding DNA sequences
  3. RNA synthesis and protein synthesis are coupled as in prokaryotes
  4. mRNA is often extensively modified before translation
  5. multiple copies of nuclear genes, and pseudogenes can occur

Ans C. RNA synthesis and protein synthesis are coupled as in prokaryotes
RNA synthesis occurs in the nucleus, but translation occurs on ribosomes in the cytoplasm. Transcription and translation are spatially and temporally separated in eukaryotic cells.

The primary distinctions between prokaryotic vs eukaryotic gene expression are as follows:

    • Eukaryotes are multicellular. Gene expression can be development and tissue specific.
    • There are multiple copies for many eukaryotic genes, and a large amount of nonessential DNA.


  • Eukaryotic genes are primarily in the nucleus. mRNAs must cross the nuclear membrane before they are translated in cytoplasm.


  • Many eukaryotic genes are interrupted by noncoding DNA, which is transcribed and then removed by a process known as splicing.
  • Eukaryotic precursor mRNA is extensively modified in the nucleus before translation.
  • Transcription and translation not coupled as in prokaryotes
  • Polycistronic mRNAs rare in eukaryotes.

Exam MCQ

The primary RNA transcript of the chicken ovalbumin gene is 7700 nucleotides long, but the mature mRNA that is translated on the ribosome is 1872 nucleotides long. This size difference occurs primarily as a result of:

  1. capping
  2. cleavage of polycistronic mRNA
  3. removal of poly A tails
  4. reverse transcription
  5. splicing

Ans E. splicing
Splicing of the pre-mRNA removes 7 introns which total 5828 nucleotides of the pre-mRNA sequence.

Exam MCQ

RNAs that catalyze biological reactions, such as self-splicing introns, are known as:

  1. enzymes
  2. spliceosomes
  3. ribozymes
  4. lariats
  5. mature RNAs

Ans C. ribozymes
A ribozyme is an RNA molecule that can catalyze a biochemical reaction.


Exam MCQ

The regions labeled A and C of the diagram are ___________________.

  1. introns
  2. snRNPs
  3. spliceosomes
  4. exons
  5. tRNAs

Ans D. exons
The 5′- and 3′-exons are separated by the intron. During the 2-step splicing reaction, the intron is removed and the exons are joined together.

Exam MCQ

The regions of DNA in a eukaryotic gene that encode a polypeptide product are called:

  1. hnRNAs

  2. exons

  3. enhancers
  4. leader sequences
  5. tRNAs

Ans B. exons
Exons are joined together during splicing, and translated on ribosomes into the polypeptide product of the gene.

Exam MCQ

mRNA will form hybrids only with the coding strand of DNA because:

  1. DNA will not reanneal at high temperatures
  2. the salt concentration will affect DNA reannealing
  3. DNA will not reanneal at low temperatures
  4. RNA:DNA hybridization follows the base-pairing rules
  5. denatured DNA will not reanneal after it is diluted


Ans D. RNA:DNA hybridization follows the base-pairing rules
An RNA molecule can form a base-paired DNA-RNA duplex molecule with a DNA that has complementary base pairing.

Exam MCQ

Which of the following is NOT involved in regulating the synthesis of RNA in the eukaryotic nucleus?

  1. active genes in euchromatin, and inactive genes in heterochromatin
  2. amplification of some genes such as rRNA genes
  3. use of different RNA polymerases to transcribe different classes of RNA D. spliceosomes that stimulate synthesis of intron-containing hnRNAs
  4. enhancers that can stimulate specific promoters


Ans D spliceosomes that stimulate synthesis of intron-containing hnRNAs

Spliceosomes are RNA-protein complexes active in processing of nuclear RNA transcripts in eukaryotic cells.

Spliceosomes are not believed to act in regulation of RNA synthesis in the nucleus.

Exam MCQ

Which of the following features would you NOT expect to find in heterogeneous nuclear RNA (hnRNA)?

  1. intron
  2. polycistronic coding
  3. polyadenylation at 3′-end
  4. 5-‘ cap structure
  5. U nucleotides

Ans B. polycistronic coding
Polycistronic mRNAs are very common in prokaryotes, but are rare in eukaryotes.

Exam MCQ

Which of the following is not part of RNA processing in eukaryotes?

  1. splicing of exons
  2. reverse transcription
  3. addition of a 5′ cap
  4. addition of a poly A tail
  5. intron removal


Ans B. reverse transcription

Reverse transcription is the enzymatic synthesis of DNA from an RNA template by the enzyme reverse transcriptase. This enzyme is not involved in the RNA processing reactions.


Transcription , false is :

a Describes the production of polypeptides from the mRNA template

b Occurs in the nucleus

c Produces single-stranded mRNA using the antisense DNA strand as a template

D Precedes 5′ capping and polyadenylation

Ans A

a False. During transcription mRNA is produced from the DNA template

b True. The mRNA product is then translocated to the cytoplasm

c True. The mRNA is complementary to the antisense strand

D True. The addition of the 5′ cap and 3′ poly(A) tail facilitates transport to the cytoplasm


  • Introns vs. exons


    • Exons contain the actual genetic information coding for protein.
    • Introns are intervening noncoding segments of DNA.
    • Different exons are frequently combined by alternative splicing to produce a larger number of unique proteins.
    • Introns are intervening sequences and stay in the nucleus, whereas exons exit and are expressed.
    • Abnormal splicing variants are implicated in oncogenesis and many genetic disorders (e.g., β-thalassemia).

Exam MCQ

A 8 year old child having β-thalassemia minor exhibits two bands on a Northern blot using a probe against exon 1 of β-globin. The smaller band is of normal size and “heavier” than the other larger band, which consists of approximately 328 additional nucleotides. One explanation for this finding is which of the following?

(A) The presence of a nonsense mutation in the DNA

(B) A mutation which creates an alternative splice site

(C) A lack of capping of the mRNA

(D) An extended poly-A tail

(E) A loss of AUG codons

4 The answer is B:

A mutation which creates an alternative splice site. The patient has developed a mutation in an intron which acts, only a small percentage of the time, as a splice donor site instead of the normal site at the intron/exon boundary. Thus, when this site is utilized by the splicesome, a piece of the intron is incorporated into the mRNA product, producing a longer than normal mRNA. This is an infrequent event, however, as judged by the fi nding that the density of the normal sized mRNA band on the gel is darker than this abnormal band. A nonsense mutation in the DNA will not affect transcription (although it does affect the protein product made from the mRNA). The lack of a cap would result in an unstable mRNA that perhaps would not be translated but would not signifi cantly change the size of the mRNA. Poly-A polymerase adds the poly-A tail and would add the same size tail to both species of mRNA.

If the polyadenylation signal were mutated, then the overall mRNA size would be larger, but there would not be two different proteins produced. Since the patient has a β-thalassemia, defective β-globin protein is being produced from the larger mRNA. Loss of methionine codons will affect translation, but not transcription.