Transcription

Question 1

Transcription

Answer 1

Copy of coding DNA into a single stranded RNA. Requires an Rna polymerase, dsDNA template, nucleotides. polymerase catalyses the formation of the phosphodiester bonds. No primer and RNA grows 5’ to 3’. RNA is direct copy of sense strand

Question 2

Prokaryotic transcription RNA polymerase

Answer 2

Two large subunits beta and beta’ (nucleotide binding and template binding respectively) and two smaller subunits (both alpha involved in enzyme assembly) and a sigma factor.

Question 3

Sigma factor

Answer 3

Sigma factors bind RNA polymerase and promoter sequences bringing them into contact to form initiation complex (covering 75-80bp either side of the start point). Different sigma factors bind to different promotors. E.g. rpoD gene produces sigma factor 70. this enables RNA polymerase to bind DNA at -50 to +20 (TTGACA -35 sequence, TATAAT -10 sequence). Acts as initiation factor staying at beginning of transcription

Question 4

Initial elongation complex

Answer 4

loses sigma, forms at ten bases and loses contact with the -35 to -55 region

Question 5

General elongation complex

Answer 5

forms at 15-20 bases and covers 30-40 bps

Question 6

Stages of transcription

Answer 6

template recognition by polymerase, unwound at promotor, initiation, elongation, termination

Question 7

Intrinsic termination

Answer 7

Intrinsic termination: The terminator sequence is usually a pallindromic sequencethat forms a stem-loophairpinstructure that leads to the dissociation of the RNApolymerase from the DNA template.

Question 8

Rho dependent termination

Answer 8

atermination factorcalledρ factor (rho factor) which is a protein to stop RNA synthesis at specific sites. This protein binds at a rho utilisation site on the nascent RNA strand and runs along the mRNA towards the RNApolymerase. A stem loop structure upstream of the terminator region pauses the RNAP, when ρ-factor reaches the RNAP, it causes RNAP to dissociate from the DNA, terminating transcription

Question 9

Classes of RNA

Answer 9

mRNA information, rRNA structural, tRNA informational/structural, RNP functional ribonucleoprotein. All transcribed from DNA and form receive post transcriptional changes

Question 10

prokaryotic mRNA

Answer 10

T1/2 for prokaryotic mRNAs a few mins. Mostly polycistronic

Question 11

RNA processing

Answer 11

bases can be modified, All stable RNAs are processed (mRNA rarely) Approx 10 nucleases involved Enzymes mainly recognise 3D structure Bases other than AGTC are formed. Three enzymes, RNase III, RNase E and RNase P are responsible for most of the primary endonucleolytic RNA processing events. The first two are proteins, while RNase P is a ribozyme (Apirion and Miczak)

Question 12

Differences in eukaryotic transcription

Answer 12

3 nuclear RNAPs., recognise DNA/Protein complex, 5’,3’ modified, mRNA processed from hnRNA, Monocistronic, Long T1/2

Question 13

RNA polymerases

Answer 13

Large enzymes with two large subunits and 10 minor units
I: nulceolus, produces rRNA, 50-70% activity, not alpha aminitin sensitive
II: nucleoplasm, nuclear RNA, 20-40%, sensitive to alpha- actinin. promotors contain TATA Box and UPEs (CAAT Box, GC Box, Octamer)
III: nucleoplasm, tRNA, 10%, species specific alpha actinin

Question 14

RNAPII CTD

Answer 14

The carboxyl end of RNA Pol II contains a stretch of seven amino acids that is RNA Pol II repeated 52 times in the mouse enzyme and 26 times in yeast. This heptapeptide has the sequence Tyr-Ser-Pro-Thr-Ser-Pro-Ser and is known as the carboxy- terminal domain or CTD. These repeats are essential for viability. The CTD sequence may be phosphorylated at the serines and some tyrosines. In vitro studies have shown that the CTD is unphosphorylated at transcription initiation, but phosphorylation occurs during transcription elongation as the RNA polymerase leaves the promoter. Since RNA Pol II catalyzes the synthesis of all of the eukaryotic protein-coding genes, it is the most important RNA poly- merase for the study of differential gene expression

Question 15

RNA polymerase II promotors

Answer 15

different combinations of TATA boxes, CAAT boxes, GC boxes and other elements. the spacing between the TATA box and the start site is important

Question 16

TFIID

Answer 16

responsible for binding to this key promoter element. The binding of TFIID to the TATA box is the earliest stage in the formation of the RNA Pol II tran- scription initiation complex. TFIID is a multiprotein complex in which only one polypeptide, TATA-binding protein (TBP) binds to the TATA box. The complex also contains other polypeptides known as TBP-associated factors (TAFIIs). It seems that in mammalian cells, TBP binds to the TATA box and is then joined by at least eight TAFIIs to form TFIID.

Question 17

TBP

Answer 17

Binding of TBP deforms the DNA so that it is bent into the inside of the saddle and unwound. This results in a kink of about 45° between the first two and last two base pairs of the 8 bp TATA element.

Question 18

TFIIA

Answer 18

TFIIA binding to TFIID prevents binding of inhibitors and allows the assembly process to continue.

Question 19

TFIIB

Answer 19

TFIIB can also bind to the RNA polymerase. FIIB acts as a bridging factor allowing recruitment of the polymerase to the complex together with a further factor, TFIIF

Question 20

TFIIF

Answer 20

Binds with the RNA polymerase II.

Question 21

TFIIH, TFIIJ, TFIIE

Answer 21

These proteins are necessary for polymerase transcription in vitro and associate with the complex in a defined order. TFIIH contains both kinase and helicase activity. Activation of TFIIH results in phosphorylation of the carboxy-terminal domain (CTD) of the RNA polymerase which allows the RNA polymerase to leave the promoter region

Question 22

mRNA capping eukaryotes

Answer 22

5-end is chemically modified by the addition of a 7-methylguanosine. addition of a GMP nucleotide to the new RNA transcript in the reverse orientation compared with the normal 3–5 linkage, giving a 5–5 triphosphate bridge. The reaction is carried out by an enzyme called mRNA guanyltransferase and there can be subsequent methylations of the sugars on the first and second transcribed nucleotides, particularly in verte- brates. The cap structure forms a barrier to 5-exonucleases and thus stabilizes the transcript

Question 23

mRNA polyA eukaryotes

Answer 23

sequence elements of a polya signal (20bp) a cleavage and addition site, and a downstream GU rich region. specific protein factors recognize these sequence and assemble a complex, cleavage takes place and then one of the factors, poly(A) polymerase (PAP), adds up to 250 A residues to the 3-end of the cleaved pre-mRNA. The poly(A) tail on pre-mRNA is thought to help stabilize the molecule since a poly(A)-binding protein binds to it which should act to resist 3-exonuclease action. In addition, the poly(A) tail may help in the translation of the mature mRNA in the cytoplasm. Not all polyadenylated (e.g. histones)

Question 24

mRNA splicing

Answer 24

The 5-end of almost all introns has the sequence 5-GU-3 and the 3-end. 5-AG-3. The AG at the 3-end is preceded by a pyrimidine-rich sequence called the polypyrimidine tract. About 10–40 residues upstream of the polypyrimi- dine tract is a sequence called the branchpoint sequence which is 5-CURAY-3. Different in yeast. the bond in front of the G at the 5-end of the intron at the so-called 5-splice site is attacked by the 2-hydroxyl group of the A residue of the branchpoint sequence to create a tailed circular molecule called a lariat and free exon 1. In the second step, cleavage at the 3-splice site occurs after the G of the AG, as the two exon sequences are joined together. The intron is released in the lariat form and is eventually degraded. Catalysed by U1 (5’ splice site), U2 (binds branch site) and U4, U5 and U6

Question 25

rRNA eukaryotes splicing

Answer 25

cleaveage in the external and internal transcribed areas gives rise to 18, 5.8, 28S rRNAs

Question 26

tRNA processing prokaryotes

Answer 26

RNases D, E, F and P once the primary transcript has folded. RNase E or Fcuts off a flanking sequence at the 3-end, at the base of a stem, to leave a precursor with nine extra nucleotides. The exonuclease RNase D then removes seven of these 3-nucleotides one at a time. RNase P can then make an endonu- cleolytic cut to produce the mature 5-end of the tRNA. In turn, this allows RNase D to trim the remaining 2 nt from the 3-end, giving the molecule the mature 3-end. Finally, the tRNA undergoes a series of base modifications.

Question 27

tRNA processing eukaryotes

Answer 27

primary transcript forms a secondary structure with characteristic stems and loops which allow endonucleases to recognize and cleave off the 5-leader and the two 3-nucleotides. A major difference between prokaryotes and eukaryotes is that, in the former, the 5-CCA-3 at the 3-end of the mature tRNAs is encoded by the genes. In eukaryotic nuclear-encoded tRNAs this is not the case. After the two 3-nucleotides have been cleaved off, the enzyme tRNA nucleotidyl transferase adds the sequence 5-CCA-3 to the 3-end to generate the mature 3-end of the tRNA. The next step is the removal of the intron, which occurs by endonucleolytic cleavage at each end of the intron followed by ligation of the half molecules of tRNA. Phosphodiesterase opens phosphate ring before folding, phosphorylation with kinase and ligation