Student Questions and Answers October 8, 2002

Transcription

1 Student Questions and Answers October 8, 2002 Q l. Is the Cα of Proline also chiral? Answer: FK: Yes, there are 4 different residues bound to this C. Only in a strictly planar molecule this would not hold, since then there is, of course, a plane of symmetry, which means that mirror images can be turned into each other. Q 2. If H 2 O::backbone hydrogen bonds are as strong as backbone::backbone ones, then why is it that there are not all backbone groups are hydrogen bonded to H 2 O, since H 2 O is smaller than the other backbone and so it would be better? Answer: AG: Hydrogen-bonding between peptide-bond groups in the backbone occurs in stretches that are buried within the hydrophobic interior of proteins, from which water molecules are normally excluded. Under physiological conditions, protein folding is assisted by chaperones that help suppress unfavourable structural conformations based on incorrect interactions among different groups. FK: Backbone backbone H-bonds may in fact be slightly more stable ( 1kcal/mole) than backbone water H-bonds (but this is not generally accepted!). Geometry (proximity) factors may be crucial. Upon formation of one H-bridge between backbone atoms the resulting local orientation of the backbone will strongly favour intramolecular formation of H-bonds rather than intermolecular ones, i.e. neighboring peptide bonds come close to their corresponding partners, and their bonding to water becomes less and less likely. Due to this high cooperativity of secondary structure formation, folding of most helices is completed on a sub-ms-scale! (β-structures usually are slower folding, since the H-bonds formed therein are long-range rather than short range interactions within the polypeptide chain, so competition with water is less easily overcome.) Q 3. How do H 2 0::backbone hydrogen bonds get turned into backbone::backbone ones in the mitochondria? Answer: AG: A special process in which H-bonding with water is replaced by internal H- bonding is not known. Proteins destined to the mitochondrial matrix are unfolded during translocation across the (inner) mitochondrial membrane, but once they enter the matrix they are assisted in their folding by chaperonins that provide a hydrophilic chamber that simulates an aqueous environment. Q 4. Some side chains have carboxylic acid- and amino groups. Can side chains form peptide bonds? If not, then why not? Answer: AG: Peptide (amide) bonds between amino acids are catalysed solely by a peptidyl transferase (a ribozyme) resident in ribosomes. The energy needed for this reaction is invested

2 by ATP hydrolysis during activation of trnas. This is not possible with side chains. FK: The only known example (to my knowledge) is in the course of blood clotting: in the final step fibrinogen is activated (by limited proteolysis) to form aggregates which to some extent are covalently linked by formation of side chain-amide bonds. Under normal conditions the reactivity of amino and carboxylic groups is insufficient without activation. Q 5. Which of the two shown structures of amino acid is the correct one; the one showing the carboxylic acid and amino groups side by side, or that in which they appear on separate sides ofthe chiral carbon? Answer: This question reflects a misunderstanding of 2-dimensional projections of 3Dstructures: The 4 residues linked to Cα are arranged at the corners of a tetraeder, so all positions are equivalent! (see below) Q 6. Van der Waals attractions; What is the specific distance, and are there any molecules that are more prone to come close enough, i.e. that are favoured to interact via v.d. W attractions? Answer: AG: At very short distances any two atoms show weak bonding due to their fluctuating electrical charges. If the 2 atoms are too close together, however, they repel each other very strongly. Two atoms will be attracted by each other until the distance between them equals the sum of their van der Waals radii. Q 7. What kinds of structures do the connecting loops have in secondary structures? \ Are they a straight line or pleated? Are there bonds or interactions between side chains or the backbone? Answer: AG: Short connecting loops very often form well defined structural elements called turns (βi-, βii-, γ, etc). FK: As a rule these sharp turns require the presence of at least one glycine and/or proline. In longer connecting loops generally only minor portions show regular secondary structure; the general outfit ranges from random coil to wide, smooth turns. These loops are attached to other parts of the polypeptide by weak interactions involving both, backbone and side chains.

3 Q 8. Hydrophobic interactions; what is the name of the cages formed by H 2 0 and what size are the liquid droplets? Answer: AG: a) clathrate structures. B) Depends of the hydrophobic solute which has to be harboured. Q 9. So there is no specific weak force except for v. d. W. attractions between hydrophobic side chains. Is the only force applied onto them represented by repulsion from the hydrophilic molecules? Answer: AG: The combined forces due to hydrophobicity and v.d.waals attractions contribute a great deal to the stability of proteins. FK: There are very rare examples of additional energetic contributions due to π-system interaction between aromatic rings of side chains if they are properly oriented (~parallel). Q 10. Are there any hydrophobic connecting loops? Answer: AG: The regular pattern of hydrogen bonds turning α-helices and β-pleated sheets into hydrophobic structures is disrupted in random coils/connecting loops. Hence, even if these consisted of hydrophobic amino acids, some hydrogen bonding potential will not be satisfied internally. This will make such structures hydrophilic, at least to some extent. FK: The general view: apolar residues buried in the hydrophobic core, polar residues at the outer shell of globular proteins is over-simplified. On the average between 30 and 40% of the solvent-accessible surface is hydrophobic, which, of course, also applies to connecting loops. Q ll. During denaturation, are all secondary structures removed, or are there some exceptions? Answer: AG: Theoretically yes. Under mild heat conditions and following chemical treatment, all H-bonds and disulfide bridges are disrupted. This leaves the protein unfolded and enzymatically inactive. However, once cooled, certain regional conformations may reappear. This holds true for epitopes required for recognition by antibodies in Western blotting that is undertaken on denatured proteins using SDS-PAGE. FK: Depends on the denaturation conditions as well as on the specific properties of the protein under investigation. As a rule, the completeness of unfolding increases in the order: high pressure < heat < detergents < urea < guanidinium chloride < perchlorates < extremes of ph; pressure and heat will leave secondary structure at least partially intact, whereas chemical denaturation usually generates random coils. Q 12. Is there a special reason why L-amino acids are prefered (asked twice)? And how is it achieved that only L-amino acids are made? Answer: AG: This preference is probably due to a quirk of evolution. Synthesis of L-amino acids involves intermediates of other metabolic pathways containing an alpha carbonyl group, and due to the nature of the reaction mechanism, the L-form is produced. FK: From the above it s clear that nature only had the choice between all-l or all-d, but you can t tell why it turned out to be all-l. Few years ago an interesting experiment was done: By chemical synthesis an analogon of HIV-protease was formed entirely from D-amino acids. It s properties turned out

4 to be undistinguishable from that of the true HIV-protease! But peptides from mixtures of D- and L-amino acids are quite unstable. Q 13. If it doesn't make a difference whether Gly or Ala are used, why then does the cell waste energy to generate alanines instead of using glycines? Answer: AG: There is a difference between Gly and Ala. But taking Leu, Ile, and Val as examples, the advantage must lie in the greater efficiency of filling up spaces in the interior of globular proteins in a very packed manner to capitalise on v.d.waals attractions. FK: Due to the greater flexibility of the peptide chain around glycines, too many glycines would destabilise proteins. On the other hand, they are found in most turns and bends, so they act as some kind of stop-codon for helices and β-sheets. Q 14. What is the biggest number of subunits of a multimeric protein? Are subunits synthesised in the same place, or do they need to be transported so as to come together? Answer: AG: The answer to this question has to do with the borderline between multimeric proteins, holoenzymes, and protein complexes. Further information will be related orally. FK: Just by chance I know that the largest known functional aggregate formed solely from polypeptides is pyruvate dehydrogenase. In eukaryotes this key enzyme is formed around a core of 60 copies of E2, containing also 22 copies of E1 and 6 of E3, forming a huge complex of about 10 megadalton size! Q 15. What is the number of amino acids in connecting loops? Answer: FK: Ranging from 4 (in turns) to > 100 in some multi-domain proteins. Q16. If right-handed α-helices are not hollow, how do left-handed helices form with the side chains inside? (Apart from left-handed helices consisting of Gly and Pro, I wanted to know with what the right-handed helix is stuffed, and is its structure tighter?) Answer: AG: a) Side-chains always face towards the outside of the helix. Neither helix is hollow (model shown). Q17. Why doesn 't the peptide bond lose its rigidness when atoms form hydrogen bonds? Answer: FK: H-bonding rather strengthens the partial double-bonding character of the peptide bond, since both the electron pull towards the carbonyl-o, and the willingness of the N to donate electrons are increased. Q18. Is there a limit to the length of a ß-pleated sheet before it folds back? Answer: AG: a) Silkworm fibroin contains long regions of antiparallel sheet, with the polypeptide chains running parallel to the fiber axis. The sheet regions comprise almost exclusively multiple repetititons of the sequence -G-A-G-A-G-S-G-A-A-G-(S-G-A-G-A-G) 8 -

5 FK: The average length of β-sheets is only 6 residues! In globular proteins the length of sec. structure elements are limited by the size of the folded protein (domain). Too large proteins are costly, folding is complicated and misfolding risk is high, and, finally, in most cases you need the protein surface for some specific property, and not the hidden core, so there is no need to make it larger than necessary! Q19. Van der Waals attractions are very weak bonds even when compared to other weak forces. And whenever atoms get close enough to establish v.d. W attractions, there are other forces like hydrophobicity which are far stronger than v.d. W attractions. So why are v. d. W. attractions of such great importance? Answer: AG: Due to their abundance! Q20. Which structure is more energetically favourable, α-helix or ß-pleated sheet? Answer: AG: These are not alternative structures. FK: This holds for virtually all proteins under (in vivo) folding conditions. In principle, helices fold faster and therefore will be formed even when an extended structure would be slightly more stable. Beta-sheets involve medium range or even long range H-bonding-interactions and therefore fold significantly later than helices, but they are intrinsically more stable, largely because H-bonds are ~ 1Å shorter than in α-helices. At elevated temperatures (or other conditions of conformational stress), this higher stability sometimes leads to helix β-strand transformations. Examples range from serpins (serine protease inhibitors) to prion protein and β-amyloid peptide. In the latter cases these newly formed strands associate with corresponding strands in other protein molecules, thus leading to characteristic protein aggregation (formation of amyloid plaques). Q 21. Which pathway (enzymes) is connected with the formation of selenocysteine? Answer: AG: Selenocysteine is a rare amino acid that is incorporated into a few proteins using the codon UGA. Normally UGA is a stop codon, but the proteins that use UGA for selenocysteine usually have mrna with significant secondary structure around the UGA. This may cause the ribosome to not recognise the UGA as stop codon, allowing the selenocysteinyl-trna to incorporate the amino acid at that point.