1
3 points Based on what you have learned about the functioning of enzymes and proteins, explain why the pH in the cell has to be buffered.
A large
... [Show More] group of enzymes show a pH optimum for their activity. This value is normally close to the pH in the cell. When the pH in the cell changes this can lead to a protonation or deprotonation of an amino acid that is important for the enzymatic activity. As a result the activity will decrease. Upon large changes in pH additional amino acids can be protonated or deprotonated, which can cause the loss of important ionic interactions and loss of secondary and tertiary structure. Most enzymes and proteins precipitate at the more extreme pH values.
2
4 points In addition to the pH, also the redox potential needs to stay constant in the cells.
(a) What species are formed that can higher the redox potential of the cytosol?
(b) What species are involved in buffering the potential and remove these species?
(c) What role does the pentose phosphate pathway play in the removal of these species?
a) The main species that can shift the potential to a higher more positive value are oxygen, superoxide, hydrogen peroxide and hydroxyl radicals. In particular the last three, the so-called reactive oxygen species (ROS) are very dangerous to the cell if their respective concentration increase too much. Iron-sulfur-cluster- containing proteins get damaged. Regulatory disulfide sites get oxidized. DNA gets damaged and eventually the cell will die.
b) Several compounds are present in the cell that can buffer the potential and/or directly remove ROS. All cells contain glutathione and thioredoxins. Plant cells also contain ascorbate. In addition, specific proteins will remove the ROS, like catalase and hydrogen peroxide.
c) Reduced glutathione (GSH) protects the cell by destroying hydroxyl free radicals. It is also the cosubstrate for glutathione peroxidase that removes hydrogen peroxide. In both cases the gluthathione gets oxidized (GSSG). Regeneration of GSH from its oxidized form (GSSG) requires the NADPH produced in the glucose 6-phosphate dehydrogenase reaction. This reaction is catalyzed by gluthathione reductase
3
4 points For a peptide with the sequence shown below determine the pI.
Glu–His–Trp–Ser–Gly–Leu–Arg–Pro–Gly
(Hint: If you cannot do that directly, first calculate the charge at different pH values and use those to find the pI region.)
pH N-term Glu His Arg C-term Net Charge
2 +H3N– –COOH >NH+ –NH3+ –COOH
+1 0 +1 +1 0 +3
5 +H3N– –COOˉ >NH+ –NH3+ –COOˉ
+1 -1 +1 +1 -1 +1
7 +H3N– –COOˉ >N –NH3+ –COOˉ
+1 -1 0 +1 -1 0
10 H2N– –COOˉ >N –NH3+ –COOˉ
0 -1 0 +1 -1 -1
Find the two ionizable groups with pKa values that “straddle” the point at which net peptide charge = 0 (here, two groups that ionize near pH 8): the amino-terminal α- amino group of Glu and the His imidazole group.
Thus, we can estimate pI = (8.0 + 6.00)/2 = 7.0
4
4 points The cyanobacterium Nostoc punctiforme can live in symbiosis with plants. The signal peptide nostopeptolide A1 plays an important role in the symbiosis. Some of the amino acids have been modified, like the methyl-proline (mPro), the acetylated Leucine (LeuAc) and there is also a butyrate group attached. Can you assign the remaining amino acids (circled with a solid line)
5
4 points Explain the principle of hydrophobic interaction chromatography.
Proteins bind to a hydrophobic ligand on the surface of a support resin under high salt concentration conditions in the mobile phase (typically >1M ammonium sulfate).
The salt promotes the hydrophobic interaction between the protein and the solid support.
To desorb the protein, the salt concentration is lowered via a decreasing salt gradient which diminishes the hydrophobic interaction.
6 For each of these purifications provide a detailed explanation whether you think the protein is pure or not and whether additional steps are needed to confirm purity. Note that the total activity is indicated, not the specific activity. [Show Less]