1 - The Structure and Function of Ribosomes

Question 1

How fast is translation?

Answer 1

Ribosomes can extend a polypeptide chain at a rate of 3-5 amino acids per second, meaning that they can produce a whole protein in a time frame that varies from minutes (100AA peptides - eg insulin, EGF) to hours (EGF receptor, titin).

Question 2

How abundant are ribosomes?

Answer 2

The ribosome is the most abundant ribonucleoprotein in the cell, but the protein elements are purely for stability and structural integrity – all the catalysis is done by rRNA ribozymes.

There are 100,000-200,000 ribosomes in an E. coli cell, meaning that they account for 25% of the bacterium’s biomass. In eukaryotes they only make up 5%.

Question 3

Where are ribosomes made?

Answer 3

The ribosome is assembled in the nucleolus

Question 4

What are ribosomes comprised of?

Answer 4

Various large subunits that are defined by their Svedberg measurement. This is so because ultracentrifugation was the first technique used to separate the ribosomal subunits.

These proteins vary between 1S and 50S, viruses having a sedimentation coefficient of 40S to 1000S.

Question 5

What affects the rate of translation?

Answer 5

Codon usage, nutrient availability, growth kinetics and energy states.

Question 6

What is the main difference between pro- and eukaryotic translation?

Answer 6

Prokaryotic ribosomes can be localised to the area of gene transcription, going as far as translating while transcription is still ongoing. In eukaryotes transcription and translation occur in separate compartments, leading to spatial and temporal separation of the two processes.

Question 7

What advantage does the separation of transcription and translation have for eukaryotes?

Answer 7

This allows for a complex layer of post-transcriptional regulation in eukaryotes not possible in prokaryotes, including more careful regulation of translation initiation, splicing control and selective degradation.

Question 8

In what kind of environment does translation occur?

Answer 8

Translation occurs in an incredibly crowded, complex and hostile environment; E. coli cytoplasm has a 300 mgml-1 protein concentration.

In eukaryotes ribosomes can be free floating or attached to the ER.

Question 9

Why are eukaryotic ribosomes thought to be more complex?

Answer 9

The increased complexity is thought to reflect their increased complexity of regulation.

Question 10

How conserved are ribosomes?

Answer 10

Even within the same kingdoms there is variability in the rRNA R-protein extensions and some possess expansion segments.

Question 11

How long and heavy are tRNAs?

Answer 11

All are 74-95 bases long and 17-20kDa.

Question 12

How many different tRNAs do eukaryotes and prokaryotes have?

Answer 12

E. coli possess 61 different tRNA, eukaryotes having around 100.

Question 13

What are rare codons?

Answer 13

All prokaryotes have a different pool of tRNA, leading to some rare codons that can slow translation. 5-8 of these in a row can cause the ribosome to stall.

Question 14

What are the structural features of a tRNA?

Answer 14

These ribonucleotides possess an acceptor stem. The 3’ overhang on the end of this is called the 3’acceptor end always ends with a CCA sequence, preceded by a single base called the discriminator base. This is involved in recognition specificity for the tRNA.

Question 15

What is the subunit composition of a prokaryotic ribosome?

Answer 15

70S complex is composed of the 50S and 30S subunits – both of which contain E, P and A sites in the peptidyl transferase centre.

Question 16

What is the weight and make-up of a prokaryotic ribosome?

Answer 16

Total of 2.5MDa, 65% of which is rRNA and the remaining 35% protein.

Question 17

Describe the prokaryotic 30S Subunit.

Answer 17

This is only 0.8MDa, possessing 1.5Mbp of mRNA. And 21 attached proteins. If the proteins are stripped away the resulting ribozyme has a sedimentation coefficient of 16S.

Question 18

What is the role of the prokaryotic 30S subunit?

Answer 18

The 30S subunit is responsible for the facilitating the mRNA tRNA interaction and therefore the accuracy of translation. It also initiates binding of the ribosome to the mRNA.

Question 19

What is the structure of the prokaryotic 50S subunit?

Answer 19

This is a 1.5MDa complex of 3Mbp of rRNA, split into two chains of 5S and 23S. These are coated in 34 proteins.

This subunit contains the peptidyl transferase centre (therefore catalysing peptide bond formation at about 20s-1) and the ribosomal exit tunnel – the hole through which the nascent chain exits.

Question 20

What is the structure of the prokaryotic ribosomal exit tunnel?

Answer 20

The ribosomal exit tunnel is lined with hydrophilic residues to prevent interaction with the new protein, making sure it remains unfolded. This tunnel is 100Å long with a diameter of 20Å at its widest point. It is lined by the 23S subunit and proteins L4 and L22.

Question 21

What are the characteristics of E. coli ribosome biogenesis?

Answer 21

E. coli ribosome biogenesis is a complex and highly regulated process, as it is incredibly costly in energy; 40% of the cell’s energy is consumed by ribosome biogenesis. In vivo this process takes only around two minutes (when at 37°C).

It is possible to reconstitute these ribosomes in vitro.

Question 22

What is the mechanism of E. coli ribosome biogenesis?

Answer 22

Biogenesis begins with transcription of the 16S, 23S and 5S rRNA as one long transcript, which is processed to yield the individual chains. Ribosomal proteins are also translated at this point and chemically modified in any ways they need to be.

The 16S and 23/5S rRNA subunits then come together to act as a framework around which the rest of the 30S and 50S subunits (respectively) can form.

Question 23

What is the history of ribosome imaging?

Answer 23

Ribosomes were identified through light microscopy as early as 1955, when they were dubbed ‘Palade particles’. They were soon renamed nucleoproteins when their nature began to be elucidated, but their structure was still a mystery. They were first called ribosomes in 1958.

The first EM micrographs of ribosomes were obtained in 1970 by James Lake, which displayed the shape of the ribosome.

Question 24

How are ribosomes imaged now?

Answer 24

Breakthroughs in structural biology, including the use of thermophilic ribosomes, and structural analysis tech has allowed for molecular detail to be obtained for ribosomes using Cryo-EM, X-Ray Crystallography and to a lesser extent NMR.

Question 25

How does EM and X-ray analysis of the ribosome compare?

Answer 25

X-Ray Crystallography have always produced internal atomic detail.
However recent advances in Cryo-EM technology now allow for information gathering on par with X-ray crystallography. It can provide dynamic information on conformational changes in the ribosome during translation and due to the binding of co-factors, which can be mapped using reconstruction.

Question 26

What structures are now available for the ribosome?

Answer 26

High resolution structural and dynamic information is now available for the 70S complex and its interaction with mRNA, tRNA, protein binding factors (eEFs etc), chaperones such as the trigger factor and substrate analogues

Question 27

What makes up the prokaryotic A, P and E sites?

Answer 27

The A, P and E sites are constucted from both of the subunits.

Question 28

Compare pro- and eukaryotic ribosomes.

Answer 28

Size - 70S(P), 80S(E)
Subunits - 50S + 30S (P), 60S + 40S (E)
Diameter (nm) - 20 (P), 25-30(E)
MW (kDa) - 2400(P), 3300(E)
RNA:Protein - 2:1 (P), 1:1 (E)
rRNA (nt) - 4500(P), 5500(E)
No. of proteins - 54(P), 80(E)

Question 29

What is the structure of the 16S rRNA?

Answer 29

The 16S rRNA of prokaryotic ribosomes possesses four domains; a 5’ domain, a central domain and a major and minor 3’ domain (also called domains I – IV, respectively). These pack tightly together into a complex helical network, providing structural support for the ribosome.

Question 30

What is the structure of the 23S rRNA?

Answer 30

The 23S rRNA is the largest structure, comprising five domains, I – V. The 5S rRNA is a far smaller and very compact subunit that binds onto the periphery of the other rRNA subunits.

Question 31

What is the PTC?

Answer 31

The peptidyl transferase centre is a ribozyme constructed from the central loop of domain V of the 23S RNA, in the 50S subunit. It is capable of a translation rate of 12-20 AA per second.

Question 32

Describe the position and structure of the ribosomal exit tunnel.

Answer 32

This goes through the 50S subunit to provide an escape for the nascent protein chain. Constructed of the 23S RNA and proteins L4 and L22 it is 100Å long (holding a 30 residue long chain) and 20Å across at its widest point. Just like you, the exit tunnel is kinked as opposed to being straight.

Question 33

That… why would you say that right now? What it wrong with you? We’re WORKING.

Answer 33

Well if you will ask the same bloody question as earlier…

Question 34

What residues make up the walls of the ribosomal exit tunnel, and what impact does this have on the nascent chain?

Answer 34

The tunnel is lined with mostly hydrophilic residues to minimise hydrophobic interaction, meaning that the protein itself is largely unfolded as it passes through the tunnel. It can however form some helical structure, which is often sensed by the tunnel. Specific protein sequences can interact with the wall to cause the ribosome to stall.

Question 35

What are the common structural properties of the ribosomal proteins?

Answer 35

Most of these have one or more surface-located globular domains, with long N or C terminal tails that are densely positively charged (Lys and Arg rich) that are buried within the rRNA structure, facilitating binding to the predominantly negative ribosome. Most of these proteins interact with multiple RNA helices.

Question 36

What protein folds are found in ribosomal proteins?

Answer 36

Many of the folds found in the globular domains resemble those found in nucleic acid binding proteins, so they recognise the ribosome through both shape and charge complementarity. For example, S2 contains a helix-turn-helix motif usually found in leucine-zipper transcription factors.

Question 37

What holds the two prokaryotic subunits together?

Answer 37

Twelve different contact points hold the 30S and 50S subunits together into the 70S ribosome. These are mainly centralised RNA-RNA interactions, with only a few RNA-protein and protein-protein interactions.

Question 38

What is the the A-minor motif and what is its role?

Answer 38

This is a very common RNA structural motif found in the ribosome, and is reliant upon the interaction of Adenosine – the most abundant base in the ribosome.

The A-minor motif is particularly important for stabilising the 23S rRNA structure. 186 of the 726 adenosine bases in the rRNA (26%) are involved in A-minor motifs.

Question 39

What is the structure of an A-minor motif?

Answer 39

This motif is characterised by an adenine inserting itself into the minor groove of an existing helix, which can cause stabilisation of helix-helix, loop-helix and loop-loop interactions. These motifs tend to cluster together.

The bonding involves the N1 and N3 adenosine nitrogens as well as its 2’-OH. There are four different types of A-minor motif, defined by the position of the 2’-OH.

Question 40

How is the 23rRNA stabilised?

Answer 40

The 23S rRNA forms a stable core similar to the hydrophobic core of a globular protein. This core consists of four regions; the head, platform, body and spur.

This structure possesses a high concentration of Mg2+ ions that stabilise it.

Question 41

What is the structure of the stalk region?

Answer 41

This is an accessory region consisting of two L7/L12 dimers, and accounts for 20% of the total ribosome protein content. Because this is a large and highly mobile region there is not yet a complete structure for it.

Question 42

What is the role of the stalk region?

Answer 42

Also known as the GTPase Associated Centre, this is a major binding site for ribosomal regulatory factors that mediate translation and the interactions that facilitate it.

Question 43

How is the stalk region attached to the ribosome?

Answer 43

The L7/12 dimers are attached to the platform region of the 23S (H43 and H44) via a binding region comprising L11 and L10. L10 binds both L7/12 dimers, producing a long ‘fishing-rod’ extension.

Question 44

What is the difference between L12 and L7?

Answer 44

L7 and L12 are identical in sequence, but L7 is acetylated where L12 is not.

Question 45

What are SecA and SecM?

Answer 45

The E. coli genes SecM and SecA are protein translocation facilitators found on the same polycistronic gene in that order.

Question 46

How is SecA inhibited?

Answer 46

SecA translation is dependent on whether or not the ribosome can bind to its Shine-Dalgarno sequence. If the ribosome pauses when transcribing SecM, as it usually does, the SecA S-D is more likely to form a stem loop that inhibits ribosome recruitment.

Question 47

How is SecA activated?

Answer 47

Under the right cellular conditions the ribosome is likely to stall at a certain point while translating SecM and cause the stem loop structure to dissolve, enabling translation of SecA.

Question 48

How does the SecM nascent chain interact with the ribosomal exit tunnel?

Answer 48

This stalling is induced by a sequence in the nascent SecM protein: F S T P V W I S Q A Q G I R A G P.

Residues within this sequence form critical interactions with the walls of the ribosomal exit tunnel, forming contacts with the L4, L22 and 23S rRNA.

Question 49

How does the ribosomal exit tunnel induce stalling?

Answer 49

The entire tunnel senses the specific sequence in the nascent chain, and although the mechanism of how this inhibits the PTC is unclear it is expected to be transmitted through the rRNA network.

Question 50

How does the ribosomal exit tunnel mediate nascent protein processing?

Answer 50

The base of the ribosomal exit tunnel acts as a hub for the binding of ribosome-associated factors that are involved in processing the nascent protein.

Question 51

What proteins mediate Protein Folding, Processing and Translocation by binding to the mouth of the ribosome exit tunnel?

Answer 51

The SRP
Processing enzymes (PDF and MAP)
Trigger Factor

Question 52

What organisms possess the SRP?

Answer 52

This does exist in both eukaryotes and prokaryotes. In E. coli the factor is known as SRP-FtsY and consists of the protein Ffh and a 4.5S RNA. However we will mostly look at the eukaryotic one.

Question 53

How does the SRP act?

Answer 53

Nascent proteins are often deposited into the ER as they are being produced (cotranslational import), so the sorting signals are always N-terminus. They usually comprise of a single positively charged residue followed by a string of 6-12 hydrophobic ones which causes elongation arrest until the SRP has done its job. Proteins without this signal remain in the cytosol

Question 54

What is recruited by a sorting signal or 6-12 hydrophobic residues? Describe its structure and function.

Answer 54

The free receptor for ER cotranslational translocation is the hexameric signal recognition particle (SRP), the only receptor or channel component that is a ribonucleoprotein. SRP is a long RNA molecule coated with six proteins that bind the nascent signal sequence and directs it and the ribosome to the Sec61 complex in the ER membrane. The SRP receptor also binds the ribosome to the membrane firmly, to prevent instability.

Question 55

What are the N-teminal processing proteins and where do they bind?

Answer 55

Peptidyl Deformylase (PDF) which is recruited by L22 and L32 and Methionine Aminopeptidase (MAP) which binds to L17/L23.

Question 56

What do PDF and MAP do?

Answer 56

PDF removes the formyl group from the N-terminal methionine, introduced by the use of the special fMet tRNA.

MAP is used to remove the entire methionine from the N-terminus.

These are sometimes used consecutively.

Question 57

What do we know about the way PDF and MAP bind to the mouth of the exit tunnel?

Answer 57

There is a large amount of competition for the emerging nascent chain - not all of these can bind at once.

Structural and biochemical crosslinking experiments show overlap in the binding sites of PDF and MAP, including regions of L17, L22, L23 and L32, so they cannot be bound simultaneously.

Hence this co-translational process occurs in a co-ordinated, sequential manner.

Question 58

Describe the structure and binding of the trigger factor.

Answer 58

The trigger factor is a 47kDa protein that binds to L23 by its N-terminal domain in a 1:1 stoichiometric ratio. Without the presence of the nascent chain it as a mean residence time of 11-15s, but with the nascent chain this increases to 50s due to the 9-30 fold increase in affinity. The trigger factor is thought to interact with hydrophobic stretches on the nascent chain.

Question 59

What is the function of the tigger factor?

Answer 59

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Question 60

*Trigger. What is the function of the trigger factor.

Answer 60

The trigger factor forms a protective cage around the mouth of the exit tunnel, through which the nascent chain threads itself. It is also known to have peptidyl-prolyl cis-trans isomerase activity as a result of part of the central domain, despite which the TF is ATP-independent.

Question 61

How accurate is translation?

Answer 61

Translation is a very high fidelity process, with a 10^-4 error rate; one mistake for every 10^4 codons translated. This is done through recognition of the appropriate tRNA anticodon for the codon.

Question 62

What is wobble base pairing?

Answer 62

where the third base in the triplet codon can recognise multiple bases – i.e. degeneracy of the 3’ base is permissible.

Question 63

What facilitates wobble base pairing?

Answer 63

the tendency of the codons for the same amino acid to differ only by the last base, and often utilises inosine to allow for more varied recognition.

Question 64

What does cognate, near-cognate and non-cognate mean?

Answer 64

When the anticodon is not complementary to the 3’ base of the codon, the match is said to be cognate.

Mismatches at positions one or two are ‘near-cognate’ and will not be accepted.

Totally different anticodons are referred to as non-cognate.

Question 65

What did biochemical studies show about how the fidelity of the codon match was sensed show?

Answer 65

That more GTP is required to stimulate the release of non-cognate tRNAs than cognate ones. The rates for GTPase activation is higher for cognate tRNAs, and they induce a conformational change in the ribosome.

Question 66

What parts of the ribosome form the decoding centre?

Answer 66

It is made up of two parts of the 16S rRNA; a guanine base at G530 of the 530 loop of helix 18 and two consecutive adenines; A-1492/1493 on helix 44.

Question 67

What conformational change occurs when a tRNA anticodon stem-loop enters the decoding centre?

Answer 67

When the anticodon stem-loop (ASL) enters the A-site it causes the formation of a ‘530 pseudoknot’ in the 16S rRNA, changing the conformation of the bases relative to the incoming.

The same change was not observed when the ribosome was imaged with paromomycin, which also binds to the A site

Question 68

How does the decoding centre interact with the ASL?

Answer 68

This 530 pseudoknot interacts with the minor groove of the mRNA/tRNA codon duplex and senses the shape of the W-C base pairs of the bases at positions one and two only.

The first and second codon position base pairs are sensed by a type I and II A-minor motif respectively, but there is not specific interaction with the third base pair

Question 69

What conformational changes occur when the correct tRNA binds?

Answer 69

The 30S subunit undergoes a large conformational change in the form of a domain closure, indicating that there is communication between this decoding centre and the rest of the ribosome.

The correct tRNA itself also changes conformation by 5° when it is bound, which may suggest that there is also communication between the tRNA and the decoding site.

Question 70

How is the prokaryotic tRNA delivered to the ribosome?

Answer 70

The prokaryotic tRNA is delivered to the A site by EF-TU*GTP.

Question 71

What does the conformational change in the 30S subunit upon binding of the correct tRNA cause?

Answer 71

Domain closure of the 30S subunit causes the shoulder region of the 16SrRNA to move to contact the EF-Tu.

Question 72

What does the interaction between the 16S rRNA shoulder region and the EF-Tu cause?

Answer 72

This shifts loops in the EF-Tu causing disruption of the Switch I structure. This exposes the EF-Tu His81, allowing it to catalyse the GTP hydrolysis.

Question 73

What does hydrolysis of the GTP in EF-TU cause?

Answer 73

A 100° conformation change in the EF-Tu that leads to its dissociation as the GDP leaves the ribosome independently.

It also causes the tRNA to shift into the A-site proper.

Question 74

What has to happen once the peptide chain has been passed to the tRNA in the A-site through formation of a new peptide bond by peptidyl transferase?

Answer 74

The tRNA and mRNA must move relative to the ribosome to place the peptidyl-tRNA in the P site and move the deacylated tRNA into the E site for dissociation. This allows for a new codon to be decoded.

Question 75

What stimulates translocation?

Answer 75

The movement is caused by a ratcheting of the two subunits – the 30S subunit rotating 6° relative to the 50S subunit.

Question 76

What does the ratcheting motion cause?

Answer 76

This shifts the deacylated tRNA into the E site, but does not move the peptidyl-tRNA into the P site.

Question 77

What moves the peptidyl-tRNA into the P site, and what else does this do?

Answer 77

This is done by another elongation factor; EF-G. The ratcheting allows this to bind, and hydrolysis of its GTP both shifts the peptidyl tRNA to the P site and reverses the ratcheting, a process that causes the EF-G to dissociate.