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Self-Assessment Short Answer Questions:
Example Answers
These answers are given in note form. Some of you submitted much fuller answers; these represent simply the minimum amount of detail we were looking for. We did specify up to 150 words!

  1. The trans configuration of the peptide bond unit is more stable than the cis form because there is less steric hindrance between adjacent side chains. Proline has a cyclic side chain (it is an imino acid) so the cis and trans configurations have almost equivalent energies.

  2. Glycine is the most flexible amino acid since it has no side chain (R = H). It can adopt positions with phi and psi angles in each of the four quadrants of the Ramachandran plot (tho' not every possible combination of phi and psi).

  3. Two Cysteine side chains can react together forming a S-S (disulphide) bond (Cystine residue), which cross-links the polypeptide chain. Disulphide bonds add stability to a protein structure; they can be intra-or inter-chain. They are rarely found in reducing environments

  4. 3.6 residues per helical turn.
    13 (not 16) - each h-bond closes a loop containing 13 atoms.
    Other types of helices: 3(10) helices (similar to alpha helices, less stable); collagen helices. A beta strand is a "helix" with 2 residues per helical turn.

  5. A protein made of L-amino acids: the stable form of the alpha helix would be left-handed. A majority of beta sheets would have left-handed twists. Allowed regions of the Ramachandran plot would have opposite signs for phi and psi.

  6. Amphipathic helix = a helix where one face is more hydrophobic than the other (i.e. it contains more hydrophobic residues). It is more often found on the edge of a globular protein - the hydrophilic face pointing towards the solvent and the hydrophobic one towards the protein core.

  7. The essential difference between type I and type II beta turns is the orientation of the peptide bond between residues i+1 and i+2. The two types have different (phi,psi) angles. In a type II turn, (phi, psi) for residue i+2 is in a normally disallowed region of the Ramachandran plot: therefore, residue i+2 is almost always glycine.

  8. Genetic code other than standard used in: mitochondria, chloroplasts and a few prokaryotes. The genetic codes of mitochondria from different types of organism are slightly different.

  9. Glu..Gln substitution is likely because: 1) only a single base change is necessary; 2) the amino acids have similar properties (size, polarity, C=O group). This substitution would lead to inactivation of the protein if that protein needed an acid group in that position for its activity.

  10. Steps involved in protein synthesis: a mRNA (messenger RNA) encoding the protein in question is synthesised by the enzyme RNA polymerase, which binds tightly to the promoter region of the DNA. Each specific amino acid is attached to its tRNA (transfer RNA) in turn and added to the growing peptide chain. The tRNA binds to the appropriate codon of the mRNA, ensuring that the amino acids are added in the correct sequence. Protein synthesis takes place on the ribosome, a large complex of RNA and proteins.

  11. Key elements in the structure of DNA: bases (Adenine Guanine, Thymine, Cytosine), deoxyribose sugars and phosphates. The bases are attached to the sugar-phosphate backbone. The DNA molecule forms a double helix through bases pairing by hydrogen-bonding; A always pairing with T and G with C. The two strands of the double helix are anti-parallel.

  12. tRNA is a cloverleaf-shaped RNA structure (L shaped in three dimensions). It has essentially four functional regions. The two most important are the anticodon loop and the 3' end at which the amino acid is attached. The anticodon loop contains a sequence complementary to the codon found in the mRNA which ensures that the correct amino acid is placed at the correct position in the protein.

  13. The destination of proteins synthesized by ribosomes obviously depends on their function. The following applies to eukaryotic cells: proteins synthesized by ribosomes on the rough endoplasmic reticulum (E.R.) may be "injected" across the plasma membrane into the lumen and thence exported (secreted by the cell) to the extracellular environment. Proteins which enter the E.R. during synthesis, and which are therefore synthesized by ribosomes on the rough E.R., have a signal peptide of 16 to 30 residues at their N-terminus. This sequence is subsequently cleaved off (by signal peptidase). The signal peptides of different proteins are not homologous, but they generally include several (4 to 12) hydrophobic residues, which are found to be essential for the polypeptide to cross the E.R. membrane. There is also a basic residue a few residues before the hydrophobic sequence.

  14. Cathepsins are lysosomal proteases involved in non-specific protein degradation. The interior of lysosomes is at about pH 5 which is close to the pH optimum of the cathepsins. They are almost inactive at cellular pH; this protects cell proteins from degradation by cathepsins if the lysosomes accidentally "leak". Some types of cathepsin are aspartic (acid) proteases and others are cysteine proteases.

  15. Protein molecules are continuously synthesised and degraded in all living organisms to ensure good housekeeping and an ability to respond to the changing needs of the cell. The concentration of individual cellular proteins is determined by a balance between the rates of synthesis and degradation, which in turn are controlled by a series of regulated biochemical mechanisms. Differences in the rates of protein synthesis and breakdown result in cellular and tissue atrophy (loss of proteins from cells) and hypertrophy (increase in protein content of cells). The degradation rates of proteins are important in determining their cellular concentrations. Protein degradation is energy dependent (requiring ATP), and is limited by the concentration of the reactants, whereas protein synthesis cannot be completed in the absence of any one of the necessary reactants.

  16. The first example of control of prokaryotic gene regulation was discovered by Francois Jacob and Jacques Monod in 1961. This was concerned with how E. coli regulated its sugar metabolism, specifically how it adjusts to growing on lactose instead of glucose. E. coli would not, ordinarily, use lactose for growth. However, if it is the only carbon source available then the bacteria will use it. Bacteria will express only those genes which are required at any particular time (ie only the ESSENTIAL RNAs and proteins). To make gene regulation as efficient as possible, prokaryotes organise genes with related functions into co-ordinately regulated units called OPERONS. The lac operon is involved in the catabolism of lactose. It is regulated by an allosteric protein which has two domains; one binds lactose, and the other has a DNA binding site. The presence of the DNA binding site is dependent on the presence of the lactose. Thus the regulatory molecule acts as a sensor to test whether lactose is present in the cell and gene expression can be modulated accordingly.

  17. An enzyme is (almost always) a protein. It is a biological catalyst, speeding up biochemical reactions while remaining unchanged itself. The enormous number of enzymes are divided into 6 classes: Oxidoreductases, Transferases, Hydrolases, Lyases, Isomerases, Ligases. Any three sensible different examples are acceptable. (Students were expected to give very brief notes about the types of reaction that their chosen enzymes catalyse.)

  18. Information in Genbank / EMBL entry: title, species & taxonomy, reference (link to Medline), link to SwissProt or TREMBL (from EMBL) if protein seq. known, database links where present, coding region & translated protein sequence where appropriate.

  19. Databases for protein sequence patterns, motifs and domains: PRINTS, PROSITE, BLOCKS, PFAM, PRODOM, SMART

  20. Transmembrane regions are typically hydrophobic as they pass through hydrophobic membranes. In most known transmembrane domains the membrane-spanning segments are alpha helices, each containing about 20 residues. It is rare for over about 8 consecutive hydrophobic residues to appear outside a transmembrane region. Therefore, calculate the "average hydrophobicity" of each segment of sequence and look for blocks of (on average) high hydrophobicity. Many different hydrophobicity scales are available.