As in the previous exercises, Swiss-PDBviewer will be used for visualizing macromolecules. The Swiss-PDBviewer User Guide contains everything you need to know about the program, so if you have a problem, this is the place where you find help.
The ribosome is one of lifes most ancient molecular machines which translates mRNA into proteins. It is remarkably conserved among organisms even in phylogenetically distinct and distant organisms. These similarities are of course greatest in areas of the ribosome that are directly engaged in the functional steps of protein synthesis. The bacterial 70S ribosome consists of two subunits (30S or the small subunit and 50S or the large subunit) that associate upon initiation of protein synthesis. It has a molecular weight of 2.5 MD of which one third is protein (roughly 50 proteins) and two thirds are ribosomal RNA, rRNA (three rRNA molecules).
S is short for Svedberg unit, the sedimentation coefficient or the rate at which a particle sediments in a gravitational field, i.e. under centrifugation. The technique of analytical ultracentrifugation was developped by Theodor Svedberg, Professor in Physical Chemistry at Uppsala University who was awarded the 1926 Nobel Price in Chemistry. Using this technique the size of the different ribosomal components were determined.
During the cycle of protein synthesis, the ribosome interacts with mRNA and tRNAs as well as protein translation factors.
The 30S subunit consists of one molecule of 16S rRNA, and 20 different proteins of various sizes. This unit is responsible for formation of the initiation complex, performs the decoding of the genetic information, and controls the fidelity of codon-anticodon interactions. The large subunit consists of 2 RNA molcules: 23S and 5S rRNA, and 34 proteins. This subunit catalyzes the reactions of peptide bond formation and peptide release. It also provides the path for the nascent polypeptide chain through a tunnel.
The first structure of a bacterial ribosome came in 2000. To date, the best resolution obtained for any of the ribosomal subunits, is for the 50S subunit from Haloalcura marismortui (an archaebacteria living in the dead sea) which was solved to 2.4 Å. Ribosomes from Thermus thermophilus, a termophile bacterium originally isolated from thermal springs in Japan, are also often used in structural studies. The ribosomes from both of these bacteria are exceptionally stable, which is an important factor when working with large complexes with many components (you don't want the complex to dissociate). In this exercise we are going to look at ribosomes, or individual subunits, from Thermus thermophilus.
Figure 1. A) Schematic drawing of a "clover-leaf formed" tRNA secondary structure with the anti-codon opposite to the amino acid attachment site. The amino acid is attached by an aminoacyl-tRNA synthetase to the 3' end of the tRNA molecule. B) and C). 3D representations of the L-shaped tRNA tertiary structure.
Which amino acid that is going to be incorporated into the growing peptide chain is determined by the codon (a triplet of bases) on the mRNA (fig 2). Many amino acids have several codon options, so each amino acid can be incorporated by a set of different codons. For example, the amino acid lysine has two codons and serine has 6 different codons. Codons for the same amino acid tend to have the same nucleotides for the two first positions and only differ in the third position. Only two amino acids, tryptophan and methionine, have one single codon. The codon for methionine is also a start signal for protein synthesis. There are also several stop codons, which are used as a signal when the protein message is ending. These are decoded not by a tRNA but by a protein called release factor.
Figure 2. The genetic code represented as a wheel. The mRNA codon is read from the center and outwards, illustrating that the first two positions of a codon are more critical for the choice of amino acid than the third one.
The reading of the mRNA codon is performed by the three bases on the tRNA that makes up the so called anti-codon. If the pairing is correct, this amino acid will be incorporated into the growing peptide chain. Since most amino acids have several codons at their disposal, this must mean that either there must be many more tRNAs than is absolutely necessary, or that some tRNAs can base pair with more than one codon. In fact, both scenarios are true. Some amino acids do have more than one tRNA but. Also, while the first two base pairs strictly have to be correctly base paired, in the third so-called wobble position of the codon, some mismatches and non-standard base pairs are tolerated. This also explains why often different codons for the same amino acid have the same bases in the first two positions.
Step 1, Decoding: An aminoacyl tRNA carrying the correct amino acid binds to the A site, delivered by the protein EF-Tu. Base pairs are formed between its anticodon and the mRNA codon positioned in the A site. The "decoding center" of the small ribosomal subunit checks that base-pairing is correct.
Step 2, Peptide bond formation: The polypeptide chain is transferred from the P-site tRNA to the free amino group of the amino acid attached to the A-site tRNA, forming a new peptide bond. This reaction is catalysed by the peptidyl transferase center of the large ribosomal subunit. This reaction is followed by a series of large conformational changes, which shift the two tRNAs into the P and E sites of the large subunit, so the tRNA which previously was in the A-site is now in the P-site and the E-site is bound with the former tRNA of the P-site.
Step 3, Translocation: Through large conformational changes catalysed by the protein EF-G, the mRNA moves by three nucleotides with respect to the the small ribosomal subunit together with the tRNAs so that the ribosome is ready to enter the next round and accept a new tRNA.These three steps are repeted in a so-called elongation cycle until the ribosome reaches a stop codon, where synthesis is stopped and the protein chain is released. Procaryotic ribosomes are remarkably efficient: within a bacterial cell, one ribosome can add 20 amino acids to a growing polypeptide chain every second.
*Choose to look at only the small subunit (2j00) by clicking off the "visible" box of 2j01
*Display only the 30S subunit by clicking off the three tRNAs (chains V, W and X), the mRNA (chain Y) and the ions if those are visible (chain W).
*Look at the proteins only by selecting the first protein chain and keep on selecting the protein chains further down in the control panel by clicking on the chains while pressing the ctrl key.
*Now do the same for the large subunit
Q1. What are the approximate proportions of protein and RNA in the two subunits (more protein, more RNA or about the same)?
*Display only the three tRNAs and the mRNA (2j00, chains V, W, X and Y)
*You will notice that only half of the A-site tRNA is visible. Only the so-called "anticodon stem-loop" (ASL) part of this tRNA (corresponding to the lower part in fig 1A) was visible in the structure, probably because the 50S subunit only holds the A-site tRNA in a fixed conformation when it carries an aminoacid or a peptide.
*Notice the kink of the mRNA between the A- and P-site codons.
Q2. What three-base codons are located in the A- P- and E-site? Which amino acids do they code for (see fig 2 above)? How many and what kind of base pairs (standard Watson-Crick or not) are formed between these codons and the three tRNA anticodons?
*Notice the kink of the mRNA between the A- and P-site codons.
*Turn on 16S RNA (2j00, chain A)
*Scientists studying the ribosome guessed that such a kink was present long before it could be experimentally verified. Let's figure out why they thought it would be there!
Q3. What is the distance between two three-base codons along a straight piece of mRNA (measure from first base in P site to first base in E site)? What is the width of a tRNA in the direction of the mRNA? What would happen if there was no kink between the A and P-site codons?
*Center on an atom in the anticodon of the A-site tRNA (use the "eye" button).
*Use the slab option under Display to remove parts of the molecule so that you can see how 16S RNA interacts with the mRNA and tRNA.
Q4. Which bases interact with the three mRNA-tRNA base pairs in the A site? What kind of interactions do they make?
Q5. Can these contacts explain why the first two base pairs have to be standard Watson-Crick while some mismatches can be tolerated in the third position?
*Turn on 16S RNA (2j00, chain A)
You can see that the ribosome is not an enzyme, but a ribozyme (catalytic RNA molecule) and no protein takes part in the reaction. The protein that comes close is L27, and it has been shown biochemically that it makes the ribosome slightly faster Probably it is helping to correctly position P-site tRNA.
The ribosome is a major target for natural and synthetic antibiotics. Detailed knowledge of antibiotic binding sites is the key to understanding the mechanisms of drug action. Structures of the ribosome or its subunits in complex with antibiotics enables a rational approach for antibiotic development and therapy strategies. Additionally, structures can be used to identify new antibiotic target sites on the bacterial ribosome.
As you have understood by now, you can often get interesting information from comparisons of a new structure with relevant structures previously solved in the lab or with structures in the PDB.
In this part of the exercise, you will look at how two antibiotics, tetracycline and pactamycine, bind to the 30S subunit. The aim is to explain how these antibiotics work, and for this, you will have to compare the antibiotic-30S structures with relevant structures of the 30S subunit without the antibiotics.
Normally, such study would start with a search for relevant structures in the PDB, but in this case we have provided you with these. The structure used for comparison will be the 30S subunit part the 70S complex with tRNAs and mRNA that you already have used (2j00). In both files containing an antibiotic, the antibiotic is colored in cyan and the mRNAs in green.
To do the comparisons, you will need to superimpose the structures using the 16S rRNA molecules (A chains).
It will be easier to see what you want if you make the 50S subunit unvisible in the 2j00-2j01 file.
*Select the 16S rRNA chain to be superimposed by clicking on chain A in the control panel (in both layers!).
*Go to "Fit/Magic fit" in the "Fit" menu and choose "backbone atoms only" and select the layers involved (choose 2j00 to be the reference).
*Compare the two structures by either manually clicking the "visible" box on and off or press ctrl and tab to toggle between the two layers.
The tetracyclines forms a group of antibiotics that have been used since the 1940s and was the first of the so-called "broad spectrum" antibiotics. They have been used extensively as anti-bacterial agents in human and veterinary medicine for decades. However, due to widespread use, resistance to this antibiotic has emerged and it can now only be used in rare cases. Research on tetracycline binding to the 30S subunit have shown that this antibiotic has one primary, functionally relevant, binding site and several secondary binding sites.
Q11. Which tetracycline binds to the functionally important site?
Q12. To which residues does tetracycline make interactions in the primary site? Are these interactions RNA sequence specific or not?
Q13. To which tRNA site (A, P or E) does tetracycline bind?
Q14. Considering where this molecule binds, can you guess its mode of action?
Q15. To which tRNA site does pactamycin bind? How do you think it inhibits translation?
Q16. Resistance to one of pactamycine or tetracycline can be obtained by mutation of the 16S rRNA molecule. Comparing the binding of the two antibiotics to the 30S subunit, can you make a guess which one it is?
Q17. Tetracycline has been used as an antibacterial agent for a long time and is quite specific for bacterial ribosomes. Pactamycine, on the other hand, is very active against ribosomes from all three kingdoms of life and is thus not so specific. Considering how these two antibiotics bind to the 30S subunit, is this what you would expect?