1. a. The pKa value of an amino acid is an indication of how strong or weak an amino acid is, based on the said acid’s dissociation constant. An amino acid that has a high pKa value is weaker than on that has a low pka value. Given that an acid’s dissociation constant is Ka, the pKa value is given by the formula: pKa = -log10Ka
pH = pKa + log ([A-]/[HA]),It follows that for Cys12, 7.4= 7.2 + log([Cys-]/[HCys]) 7.4-7.2 = log([Cys-]/[HCys]) 0.2 = log([Cys-]/[HCys]) 100.2= Cys-/HCys Therefore, Cys-/HCys = 10-4Similarly, for Cys15, 7.4 = 5.1 + log([Cys-]/[HCys]) 7.4 – 5.1 = log([Cys-]/[HCys]) 2.3 = log([Cys-]/[HCys]) 102.3 = Cys-/HCys Therefore, Cys-/HCys = 102.3
Generally, the calculated values could be lower than typical values because in the physiological state, the presence of water adds to the amount (and thus the concentration) of hydrogen ions available in solution.
2. All the amino acids mentioned have hydrophobic side chains, which when closely apposed to each other get attracted to one another through van der Waal’s forces. This attraction contributes to the folding such that the presence of side chains adds to the surface are to volume ratio of the amino acids that have these moieties. This surface area to volume ratio is drastically reduced when the said moieties are absent, and the van der Waal forces of attraction are also relatively weaker, hence the amino acids lacking these moieties assume comparatively straighter structures.
3. STQVYGQDVWLPAETLDLIR is found in isocitrate dehydrogenase, which is a protein found in the Kreb’s cycle. Isocitrate dehydrogenase is used to remove hydrogen ions from isocitrate during Kreb’s cycle, and it acts as a catalyst in the process. One other substrate of the enzyme is alpha ketoglutarate, and when it is in the process of losing a carbon atom, the intermediate in the reaction is oxalosuccinic acid.
5. a. Fluroescence technique should be used, because other techniques would change the structure of the protein, causing the objective not to be met with regard to maintaining the structure so required for analysis.
b. Circular dichroism should be used so that each domain is colored differently from the other to enable visualization.
c. Nuclear magnetic resonance should be used when determining whether a protein becomes more stable with its bound ligand than when it is not bound. This is a vibrational assay, and when the vibration is altered the sample being analyzed can disintegrate, showing that it is unstable at a certain frequency of vibration. In the case of such a protein binding to the ligand, the vibrational frequency at which the unbound protein disintegrates, and that at which the bound protein does the same can be compared to determine the level of stability. The sample that requires higher frequencies to disintegrate is more stable than that which disintegrates at a lower frequency.
d. Electron paramagnetic resonance should be used when resolution is required at the Angstrom level. An Angstrom is a relatively small measurement that can be used to determine the distance between two electros or between an electron and the nucleus. Whereas other methods of determining protein structure do not allow for such resolution, electron paramagnetic resonance does, and is the best fit for analyzing how the active site of an enzyme is oriented with respect to its side chains when a ligand has been bound and when it is devoid of a ligand.
e. x-ray crystallography should be used when one intends to determine whether a protein associates with lipid bi-layers. Proteins can be frozen to elicit the crystalline structures reminiscent of solid ice. In this form, their activity is arrested, and they lend themselves to analysis which can only be carried out through x- crystallography.
6. Such alignments can give information about organisms which are closely related based on the proteins their share in common. They also give information concerning the lengths of the polypeptide chains required for the proteins to function, with the additional information about how long an amino acid sequence should be for the protein to function. They also give information regarding protein-protein interactions and gene-gene interactions such that when one protein is being expressed, a person can tell which other proteins are going through a similar process, and which ones have been switched off.
b. The sequence alignments can reveal whether proteins shall have structural and functional similarities, because they show the similarities in gene sequences as well as any differences if they are there. Similarities in sequences of genes, as well as similarities in their lengths portend similarities in the proteins being expressed by different organisms.
c. Some amino acids are known to easily mutate, while others rarely do so. The amino acids best used for analyzing mutations at proteins’ active sites are the ones which are most likely to undergo mutations, rather than those which would not be expected to do so. The former are thus given the priority over the latter. Since these mutations are most likely to occur among amino acids which contain two similar bases, such are the amino acids to be targeted. When two similar bases are present, it is easier for a third base to be substituted for the dissimilar base to make a nonsense codon.
d. I would differ with such a suggestion, because while processing the sample, there are many chemical processes involved, which affect the integrity of the proteins present in it. Trying to determine whether the structure of the wild type and that of the sample being analyzed are similar would be a self-defeating quest, because it is already expected that they shall not have similar secondary structures. Processing of the sample interferes with the intermolecular bonds within the proteins, ensuring that the structure is not maintained as it would be under normal physiological conditions. Once the primary structure has been disrupted in this manner, it is not likely that the secondary structure shall remain the same, nor can it be reverted to look as it did before the processing had begun.
7. a. The protein with a larger mass shall be retained in preference to the one with the smaller mass, because as the masses increase so do the coulombic forces of attraction between protein molecules. Therefore, the protein with pl of 7.7 (protein B) shall remain in the ion exchange matrix, while protein A shall be eluted out of the matrix.
b. The protein with the least sized particles shall be eluted the most. Therefore, one should look at the elution volumes to determine which protein had an elution volume that was neither too high nor too low, given that when the molecular weights are compared, 50kDa lies somewhat midway between the other two.
c. The two bands so observed represent a protein such as the cholera toxin or the ribosomes, which are composed of two parts; one larger than the other. The protein might weigh 50kDa in total, but through SDS-PAGE, it gets disintegrated into its constituent parts, whose individual masses do not add up to 50kDa.
d. Ribosomes may be collected and assayed through a size-exclusion chromatography channel before being passed through SDS-PAGE, and analysis made of the bands produced to see whether they correspond to a single band or two bands will be formed according to literature.
8. a. The ability of a protein to bind to its corresponding ligand depends on the molecular structure of the ligand, the hydrophobicity of the protein involved, and the presence of charges around and within the binding site. A change in the molecular structure of either the protein or the ligand to be bound makes binding impossible due to lack of conformity, and without opposite charges within the ligand binding site or around it, there will be no attraction between the protein and its ligand, hence no binding. In addition, hydrophobic proteins shall attract hydrophobic ligands more than they shall attract hydrophilic ligands, and the reverse is also true.
b. In this case, the association constant may be likened to the dissociation constant of acids, and the other variables compared to the pH, concentration of the free ions, and concentration of the acid. Hence, Ka =10(-pKa), and
assuming the ligand to be the positively charged moiety,
pL = -log[Ka] + log([P]free/[PL])
c. Ka depends on Kd, and when Ka is high, the attraction between the protein and its ligand is much stronger, meaning that the dissociation constant is low. The relationship between these two measures is one of inverse logarithmic proportionality such that a linear increase in one value leads to the logarithmic decrease of the other variable. In other words, when there is a decrease in one value, the other variable experiences an exponential increase. Therefore, Ka = -log10 Kd.
d. With a decrease in the binding free energy, through the loss of a hydrogen bond, there is a corresponding decrease in the energy of association. The energy required for the ligand and the protein to bind in this case will be much higher than that which is currently available, and the complex formed between the protein and its ligand cannot be maintained in the bound state for as long as it would be maintained otherwise. The most potent of the three inhibitors is the one that has the least constant of dissociation, implying that it has the largest constant of association. In this regard, inhibitor C is more potent than the other two inhibitors.
9. a. Integral proteins composed of beta barrels have hydrophobic amino acids concentrated on surfaces adjacent to lipids while alpha helices show a diffuse distribution of such amino acids, without a bias for either surface. Aromatic rings may be found around the interfaces of the lipid and cytoplasmic layers.
b. Proteins are products of hydrocarbon chemistry, implying that they exist in the form of chains that can either be simply straight in structure, or they can be branched chains. Two proteins can have the same number of carbon atoms, in this case a 20 residue hydrophobic segment in one, but the other still having the same 20 residues fails to be hydrophobic because the one is a straight chain and the other is a branched chain. Branched chain proteins will have greater hydrophobicity tendencies than those that are not branched, and this difference in structure explains the observable difference in hydrophobic properties.
c. It is these proteins that contain charged residues, which enable them to be oriented in a manner that allows the lipid bilayer to maintain its integrity. This characteristic would serve no purpose in the free proteins.
10. a. higher affinity for the substrate implies that there will be longer effect. The mutation causes increased polarity within the binding site region.
b. Hydrolysis is GTP-dependent, and when G protein binds GTP, GTP is not available for the hydrolysis reaction. Longer/stronger binding implies that the GTP is held much longer than usual hence slowing down the process of hydrolysis.
c. When the bonds between hemoglobin subunits weaken, the efficiency of the whole molecule in binding oxygen molecules is affected. In the mutant state, hemoglobin will bind oxygen with lower affinity because its structure has been modified by the mutation.
d. An increase in pore size facilitates the transportation of potassium ions across the membrane. More ions are thus pumped across than when the pore size remains normal.
e. The sulphydryl moiety is necessary for binding to occur, and this happens best when this moiety is exposed. The mutation causes a change in the structure of the protein, rendering the sulphydryl position inaccessible to insulin.
References
Fleming, S., & Ulijn, R. V. (2014). Design of nanostructures based on aromatic peptide amphiphiles. Chemical Society Reviews , 8150-8177.
Siciliano, C. (2013). Quantitative determination of fatty acid chain composition in pork meat products by high resolution 1H NMR spectroscopy. Food Chemistry , 546–554.
Vogelstein, B. (2014). Genetic alterations in isocitrate dehydrogenase and other genes in malignant glioma.
