Translation

Question 1

Translation

Answer 1

the translation of mRNA to a protein. in eukaryotes translation and transcription can be coupled.

Question 2

tRNA features

Answer 2

must have: Anticodon loop – if mRNA is 5’AUG then anticodon is 5’CAU. CCA – top three circles on diagram in dark grey where amino acid is loaded. Amino acyl synthetases – enzymes that add amino acids to correct tRNA

Question 3

tRNA structure

Answer 3

modified bases e.g. dihydouridine. Clover structure folded based on 9 h bonds to give tertiary L shape with D loop, acceptor sten, tpsiCG loop, variable loop and anticodon loop.

Question 4

amino acyl trna synthetase

Answer 4

fall into one of four classes of subunit structure, being either alpha, alpha2, alpha4 or alpha2beta2. The polypeptide chains range from 334 to over 1000 amino acids in length, and these enzymes contact the tRNA on the underside (in the angle) of the L-shape. They have a separate amino acid-binding site. The synthetases have to be able to distin- guish between about 40 similarly shaped, but different, tRNA molecules in cells, and they use particular parts of the tRNA molecules, called identity elements, to be able to do this

Question 5

amino acylation of tRNAs

Answer 5

AMP is linked to the carboxyl group of the amino acid giving a high-energy intermediate called an aminoacyl adeny- late. The hydrolysis of the pyrophosphate released (to two molecules of inorganic phosphate) drives the reaction forward. In the second step, the aminoacyl adenylate reacts with the appropriate uncharged tRNA to give the aminoacyl-tRNA and AMP. Some synthetases join the amino acid to the 2- hydroxyl of the ribose and some to the 3-hydroxyl, but once joined the two species can interconvert. The formation of an aminoacyl-tRNA helps to drive protein synthesis as the aminoacyl-tRNA bond is of a higher energy than a peptide bond and thus peptide bond formation is a favorable reaction once this energy-consuming step has been performed.

Question 6

Proof reading

Answer 6

Binding to incorrect amino acid  conformational change in enzyme which causes hydrolysis of incorrect aminoacyl-adenylate or by transfer of the amino acid to tRNA followed by hydrolysis. low error rate (1 in 50000 codons)

Question 7

Codon wobble

Answer 7

5-anticodon base was able to undergo more movement than the other two bases and could thus form nonstandard base pairs

Question 8

Selection of AUG in prokaryotes

Answer 8

Shine dalgarno sequence (7+-2 from AUG codon). 16rRNA commplementary to the SDS in small ribosome subunit. Brings ribosome into contact with first AUG. Polycistronic so multiple AUGs

Question 9

fMET-tRNA(f)

Answer 9

has unique features which distinguish it for initiation:
Formyl methionine
No base pairing at 5’
3GC pairs
Lacks alkylated adenosine in anticodon loops (allows for 2 start codons GUG and AUG

Question 10

Selection of AUG in eukaryotes

Answer 10

monocistronic, so only one AUG selected

Question 11

Kozak’s scanning hypothesis

Answer 11

Ribosomes bind and then scan until strong consensus sequence: G/AXXAUGG strong, GXXAUGA moderate, GXXAUGU/C weak, U/CXXAUGU/C inactive

Question 12

Ribsomes prokaryotes

Answer 12

50s (31 proteins, 23s and 5s RNA) and 30s (21 proteins 16sRNA)

Question 13

Ribosomes eukaryotes

Answer 13

60s (45 proteins, 28, 5.8 and 5sRNA) and 40s (30 proteins, 18sRNA)

Question 14

Prokaryotes IF1

Answer 14

associates with the30Sribosomal subunit in the A site and prevents anaminoacyl-tRNAfrom entering. It modulates IF2 binding to the ribosome by increasing its affinity. It may also prevent the50Ssubunit from binding, stopping the formation of the 70S subunit. It also contains a β-domain fold common for nucleic acid binding proteins. binds to 30s with IF3.

Question 15

Prokaryotes IF2

Answer 15

binds to aninitiator tRNAand controls the entry of that tRNA into the ribosome. IF2, bound to GTP, binds to the30SP site. After associating with the 30S subunit, fMet-tRNAfbinds to the IF2 then IF2 transfers the tRNA into the partial P site. When the 50S subunit joins, ithydrolyzesGTPtoGDPandPi, causing a conformational change in the IF2 that causes IF2 to release and allow the 70S subunit to form

Question 16

prokaryotes IF3

Answer 16

is required for the30Ssubunit to bind to the initiation site inmRNA. In addition, it has several other jobs including the stabilization of free 30S subunits, enables 30S subunits to bind to mRNA and checks for accuracy against the firstaminoacyl-tRNA. It also allows for rapid codon-anticodon pairing for the initiatortRNAto bind quickly to. IF3 is required by the small subunit to form initiation complexes, but has to be released to allow the50Ssubunit to bind

Question 17

steps in initiation prokaryotes

Answer 17

IF1 and IF3 bind to a free 30S subunit. This helps to prevent a large subunit binding to it without an mRNA molecule and forming an inactive ribosome. ● IF2 complexed with GTP then binds to the small subunit. It will assist the charged initiator tRNA to bind. ● The 30S subunit attaches to an mRNA molecule making use of the ribosome- binding site (RBS) on the mRNA (see Topic Q1). ● The initiator tRNA can then bind to the complex by base pairing of its anti- codon with the AUG codon on the mRNA. At this point, IF3 can be released, as its roles in keeping the subunits apart and helping the mRNA to bind are complete. This complex is called the 30S initiation complex. ● The 50S subunit can now bind, which displaces IF1 and IF2, and the GTP is hydrolyzed in this energy-consuming step. The complex formed at the end of the initiation phase is called the 70S initiation complex.

Question 18

elongation prokaryotes EF-tu

Answer 18

EF-Tu is required to deliver the aminoacyl- tRNA to the A-site and energy is consumed in this step by the hydrolysis of GTP.

Question 19

elongation prokaryotes EF-ts

Answer 19

EF-TuGDP complex is regenerated with the help of EF-Ts. In the EF-Tu–EF-Ts exchange cycle

Question 20

elongation prokaryotes peptidyl transferase

Answer 20

peptidyl transferase activity of the 50S subunit can now form a peptide bond between these two amino acids without the input of any more energy, since energy in the form of ATP was used to charge the tRNA

Question 21

elongation prokaryotes ef-g

Answer 21

EF-G (translocase) and GTP binds to the ribo- some and, in an energy-consuming step, the discharged tRNA is ejected from the P-site, the peptidyl-tRNA is moved from the A-site to the P-site and the mRNA moves by one codon relative to the ribosome. GDP and EF-G are released

Question 22

prokaryotes elongation steps steps

Answer 22

aminoacyl tRNA delivery, petide bond formation, translocation of tRNA from P site to E site

Question 23

termination prokaryotes

Answer 23

RF1 recognizes the codons UAA and UAG, and RF2 recognizes UAA and UGA. RF3 (bound to GTP) helps either RF1 or RF2 to carry out the reaction.The release factors make peptidyl transferase transfer the polypeptide to water rather than to the usual aminoacyl-tRNA, and thus the new protein is released. To remove the uncharged tRNA from the P-site and release the mRNA, EF-G together with ribosome release factor are needed for the complete dissociation of the subunits. IF3 can now bind the small subunit to prevent inactive 70S ribosomes forming. Leaves seperated mRNA, small and large subunits, release factor and tRNA

Question 24

initiation eukaryotes

Answer 24

There are at least 12 reasonably well defined initiation factors involved in eukaryotic protein synthesis, and some have analogous functions to the three prokaryotic IFs

Question 25

difference between eukaryotic and prokaryotic protein synthesis

Answer 25

more intiation factors for eukaryotes, eukaryotes: Control by phosphorylation of eIF2 alpha subunit. One termination factor. Transcription/translation not coupled. T1/2 is usually long

Question 26

steps in eukaryotic synthesis

Answer 26

initiator tRNA joins to make a complex of three components (ternary complex), the initiator tRNA, eIF2 and GTP. multifactor complex then formed eIF1, eIF2-GTP-tRNAi, eIF3 and eIF5. Binds to 40s subunit via eLF1a to form 43s pre-initiation complex. mRNA interacts with eIF4B and eIF4F using up ATP and being unwound by eIF4A. 43S pre-initiation complex binds to the mRNA complex via the interactions between eIF4G and eIF. ATP used to scan for AUG. eIF5B displaces eIF1, eIF2, eIF3 and eIF5 and GTP is hydrolyzed. eLF2 enters cycle to regain GTP. eIF1A and eIF5B are released when the latter has assisted 60S subunit binding to form the complete 80S initiation complex.