Module 04 - mRNA Processing

Question 1

What are the differences in mRNA processing between Prokaryotes and Eukaryotes?

Answer 1

Prokaryote: mRNA can be directly transcribed, sometimes while still being synthesized and directly translated by ribosomes - No compartmentalization
Eukaryote: Transcription and pre-mRNA processing takes place in nucleus, and the mature mRNA has to exit the nucleus to the cytoplasm to be translated

Question 2

What are the 3 important structural feature of the mature mRNA in Eukaryotes?

Answer 2

(1) 5’ cap
(2) 3’ Poly(A) tail
(3) Coding region composed of the exons spliced together with no introns included which will be translated into a protein

Question 3

What is contained in the 5’ and 3’ untranslated regions (UTRs)?

Answer 3

Contain important regulatory regions

Question 4

What is the structure of the 5’ caps ? (3 things)

Answer 4

(1) 5’ cap is a residue of 7-methylguanosine (7-meG).
(2) 7-methyguanosine (7-meG)) is linked to the 5’-terminal residue through an unusual 5’,5’ triphosphate linkage by guanylyltransferase through condensation of a GTP molecule
(3) Other Methyl groups are often added at the 2’ hydroxyl of the ribose sugars at first or second nucleotide adjacent to the cap

Question 5

What mediates 5’ capping of mRNA

Answer 5

Guanylyltransferase enzyme

Question 6

What are the function of 5’ cap?

Answer 6

Prevents recognition of the 5’ by exoribonucleases and mediates binding of mRNA to ribosome to facilitate translation

Question 7

What are the steps in the addition of the 3’ polyadenylation tail? (3)

Answer 7

(1) mRNAs contain a polyadenylation (poly(A)) signal embedded in the sequence transcribed
(2) RNA Pol transcribes the poly(A) site at the terminal end of the transcript
Cleave sites marked by 2 sequences elements:
-5’-AAUAA located 10 to 30 nucleotides on the 5’ side of the cleavage site
- Less well-defined sequence rich in G and U residues located 20 to 40 nucleotides downstream of from the cleavage site
(3) Polyadenylation factors bind the poly (A) signal, initiating mRNA cleavage.
(4) “A” residues are added to the free 3’-OH end of the mRNA in a poly(A) polymerase PAP-catalyzed reaction. These ‘A’ residues are bound by PABPs to protect the tail from degradation from exoribonucleases

Question 8

What is the main function of Pol(A) tails?

Answer 8

To protect the 3’end from nucleolytic degradation

Question 9

What are the 2 elements of the mRNA cleavage site?

Answer 9

• 5’-AAUAA located 10 to 30 nucleotides on the 5’ side of the cleavage site
• Less well-defined sequence rich in G and U residues located 20 to 40 nucleotides downstream of from the cleavage site

Question 10

What are the 3 pre-mRNA modification?

Answer 10

(1) addition of 5’ cap
(2) Addition of poly(A) tail
(3) Splicing

Question 11

What is splicing?

Answer 11

Introns in primary RNA transcripts are excised and the exons are joined to form the mature RNA

Question 12

What modification represents a major modification in mRNA processing between bacteria and eukaryotes?

Answer 12

Splicing, because prokaryote genes do not contain introns

Question 13

Do higher eukaryotes, like humans, have more introns or exons?

Answer 13

Introns

Question 14

Why do eukaryotic genes have introns if they are just spliced out?

Answer 14

Provides much more flexibility in term of the proteins that can be generated from a specific sequence

Question 15

What is alternative splicing?

Answer 15

A process in which exons in the primary transcript from a single gene are spliced together in several combinations to produce different mRNAs and thus different polypeptides
- selected exons can vary, but order does not

Question 16

What is a splice site?

Answer 16

nucleotide sequences within the intron and at the border between introns and exons

Question 17

What is the main benefit that alternative splicing offers to an organism?

Answer 17

Allows for a greater diversity of gene products as one gene can give rise to multiple different protein products

Question 18

What is the main consideration for qPCR and why?

Answer 18

Primer design, because you want primers that will amplify a specific cDNA but no any gDNA

Question 19

How can you design a qPCR primer to tonly amplify cDNA and not gDNA?

Answer 19

Design your primers to bind to two exonic regions (that span a large intron) then the polymerase will not be able to amplify the large region of gDNA

Question 20

In reverse transcription how can you make sure your cDNA only contains RNA?

Answer 20

By using a poly (T) sequence as a primer to make use of the 3’ poly(A) – called oligo dT primers