Lecture 5-Transcription II (Martin)

Question 1

describe the 5’ cap

Answer 1

• triphosphate bridge
• 7-methyl guanosine
• possibly methylation at some of the 2’ OHs that are 3’ to the cap to prevent nucleophilic attack

Question 2

What are the functions of the 5’ cap (4)

Answer 2

• causes efficient splicing
• allows it to leave the nucleus efficiently
• allows efficient transcription
• increases the T1/2 be preventing 2’OHs from attack

Question 3

Explain how the cap is added to the mRNA

Answer 3

• PIC phosphorylated by IIH and multiple nuclear proteins will bind to the phosphorylated sites including the Cap synthesizing protein
• This forms the cap structures then dissociates
• Cap binding Complex (CBC) then binds here and holds the mRNA here

Question 4

What is special about the lariat method of self splicing?

Answer 4

• the adenosine makes 3 bonds at the corner of the lariat
• 2’ OH of adenosine in intron acts as the nucleophile
• A will eventually form 3 bonds at the same time

Question 5

snRNPs all contain ______

Answer 5

RNA

Question 6

What is the order of snRNP binding to the RNA?

Answer 6

• U1, U2

- U4/U6, U5

Question 7

What is the order of release of snRNPs from the RNA?

Answer 7

• U1, U4

- U2, U5, U6

Question 8

U1 binds where?

Answer 8

5’ end of intron

Question 9

U2 binds where? Whats special about this?

Answer 9

branch point (3’ of intron) but has a mismatched A that will act as a nucleophile

Question 10

What is the inactive splicesome?

Answer 10

• U1, U2, U4, U5, U6

Question 11

What is the active splicesome?

Answer 11

• U2, U5, U6

Question 12

Eukaryotic splicing requires ______

Answer 12

ATP

Question 13

Eukaryotic splicing will eventually form what structure?

Answer 13

• lariat

Question 14

Where do eukaryotic snRNPs bind?

Answer 14

• phosphorylated sites of the pol

Question 15

Why is the location of snRNPs important?

Answer 15

• it allows for ordered splicing of the transcript so exons aren’t paired incorrectly

Question 16

Beta thalessemia is the result of _____

Answer 16

• beta subunit being synthesized more slowly than the alpha subunit because of a point mutation in the branch region which causes reduced binding of U2 to the branch point and inefficient splicing
• can sometimes lead to abnormal or truncated protein

Question 17

Explain how eukaryotic polymerases terminate transcription (4)

Answer 17

• AAUAAA sequence transcribed at end of transcript
• endonuclease cleaves transcript nearby (30bp downstream)
• polyadenylate pol (ATP requiring) makes the polyA tail
• polyA tail wraps around polyA binding protein to leave nucleus

Question 18

What are the functions of PolyA binding protein (3)

Answer 18

• helps it leave the nucleus
• improves translation efficacy in cytoplasm because it associates with 5’ cap
• protects from nucleases

Question 19

What happens if you shorten the polyA tail?

Answer 19

• you make the RNA more susceptible to nucleases and therefore shorten the half life

Question 20

How can the cell regulate the amount of protein in a cell through the poly A tail?

Answer 20

• if a gene has two polyA sites then the cell controls which site it uses
• at the 3’ unTRANSLATED end of the transcript are factors that increase the stability of it and therefore give it a longer half life
• if you splice at the earlier site then you decrease the amount of mRNA in the cell

Question 21

Whats an example of something that uses various polyA splice sites to regulate protein concentration?

Answer 21

• calcintonin

- CGRP (Calcitonin gene related protein)

Question 22

Explain the difference between HPV infected cells and tumorigenic HPV infected cells.

Answer 22

• HPV infected cells have not integrated the viral DNA into their genome so the transcripts have short T1/2s and fewer proteins made
• oncogenic virus is incorporated into the chromosome, loses the 3’ end of the transcript and thus is more stable

Question 23

Explain the synthesis of rRNAs (3)

Answer 23

• all on one transcript
• after trasncription they are methylated on 2’OH groups to prevent nucleophilic attack
• cleaved

Question 24

RnaseP

Answer 24

cuts 5’ end of tRNA

Question 25

RnaseD

Answer 25

cuts 3’ end of tRNA where CCA sequence will be added

Question 26

The D arm is rich in _____

Answer 26

• dihydrouridine

Question 27

T arm is rich in ______

Answer 27

• ribothymidine

- pseudouridine