BCHE 5180/6180: Spring Final (Graded A+)

BCHE 5180/6180: Spring Final (Graded A+) BCHE 5180/6180: Spring Final (Graded A+) 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 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. a 2 points Purification A Total Activity Step 1 5000 Step 2 4000 Step 3 2000 Step 4 1500 There is only one band on the SDS-PAGE so this protein is pure. b 2 points Purification B Total Activity Step 1 2000 Step 2 800 Step 3 600 Step 4 250 The protein might be pure after step 4 in the case that it is a protein with 3 subunits, but we will not know until more purification steps are performed. A size-exclusion step should be the next step to see if the three bands can be separated or not. In this case the total activity does not change and the SDS-PAGE shows an identical pattern of after step 3 and step 4, which indicates that the protein is pure and it has 2 subunits. 7 4 points In question 6, only the total activity was indicated and not the specific activity. (a) Explain how these are different. (b) Would your answers have been different if the specific activity would have been given for each step? The total activity might go down in a step due to loss of the protein that needs to be purified or just because the protein is not stable and loses activity over time. That is why the specific activity is normally calculated. This is the total activity divide by the total amount of protein present. This number should go up when you remove unwanted protein since you get more and more pure protein. When the specific activity in each step increases these steps and the purification are successful. If the activity goes down something went wrong. If the activity levels out it means the protein is probably pure but this should always be confirmed with another method like SDS-PAGE. The answers to 6 should not have been different. In 6a the protein can be considered pure just based on the SDS-PAGE. In 6b it could be pure but more steps are needed. In 6c both the total and the specific activity should have leveled out and the same conclusion should be reached. 8 6 points (a) Describe/draw in detail the ‘catalytic cycle’ of the folding of proteins by the GroEL/GroES complex. Explain in detail the different steps. (b) Why does this process require ATP? A folded protein has a much lower potential energy than the unfolded proteins and therefore the folding process should in principle produce energy not require energy. a) • The binding of the substrate protein to the rim of the cavity on one ring of the GroEL complex is considered the first step in the folding cycle. • This causes a conformational change allowing ATP to bind to the equatorial domains of the same ring • Now GroES can bind to the cis-ring and the substrate protein is transferred into the ring cavity • This is followed by hydrolysis of ATP, which causes the conformational change in the ring proteins exposing the hydrophilic surfaces • Now the substrate proteins is forced to (partially) fold. • A new substrate molecule can now bind to the trans-ring, followed by the binding of ATP. • This causes the release of folded products and ADP are released from cis-ring • A new reaction cycle starts. b) The most important aspect is that each step does not become a dead end. Therefore there is a continuous cycle of events where each step automatically result in the next step. The use of ATP is part of this process. The binding of ATP and release of ADP are all signals for the next step in the cycle, while the hydrolysis is important to provide energy for a large conformational change in the ring subunits. 9 6 points What is the effect of O2, H+, CO2, and 2,3-bisphosphoglycerol (BPG) on Hemoglobin Oxygen is an allosteric regulator of hemoglobin and is also the transported ligand. Hemoglobin is a protein that can be present in two different states, the T-state with low affinity for molecular oxygen and the R-state with high affinity for molecular oxygen. The binding of oxygen to the T-state cause a conformation change causing the protein to adapt the R-state. H+ binds to specific amino acids like His HC3 when the pH is the blood is lower. These amino acid can now form ion bridges with other amino acids. These bridges, however, are only formed in the T-state. As a result the equilibrium between the R- and T-states shift towards the T-state at low pH and hemoglobin binds oxygen less tightly releasing most of the oxygen bound. The CO2 produced in the tissues is the main reason the pH is lowered due to the formation of bicarbonate and protons upon reaction with H2O. The CO2 itself will react with the N-terminal ends of the four subunits and form a carbamate. The carbamate can also form a hydrogen bond with another group but only in the T-state. The effect of both H+ and CO2 is called the Bohr Effect. BPG is an allosteric modifier that binds to the central cavity of hemoglobin. BPG affects hemoglobin affinity for oxygen by binding at a site different than the oxygen binding site and promoting a shape change in the molecule. BPG fits in a cavity between the four subunits when Hemoglobin is in the T-state. Therefore it affects the equilibrium with O2. The result of BPG binding is that the oxygen binding curves shift right. The presence and effect of BPG is the main reason hemoglobin is an allosteric protein. 10 3 points Explain how the effects of sickle cell disease demonstrate that hemoglobin undergoes a conformational change upon releasing oxygen. In Hemoglobin S, the wild-type glutamate at residue 6 of the B-chain is replaced by valine. When oxygen is bound, both Hemoglobin A and Hemoglobin S are soluble, but in the deoxy-form. Hemoglobin S (but not Hemoglobin A) becomes very insoluble, due to exposure of the hydrophobic valine residue. This exposed “patch” causes aggregation of deoxy-Hemoglobin S into long insoluble fibrous aggregates, resulting in distorted shapes of the red blood cells (and leading to the symptoms of the disease). 11 6 points Write out the Michaelis-Menten equation and the reaction equation that describes the mechanism for enzyme action used as a model by Michaelis and Menten. Under what assumptions and conditions can we use this equation? (List all 5). k1  k3  V [A] E  A  EA  E  P V  max 0 K [A] k2 k4 M 1) [E] is kept constant. 2) There is an EA complex formed and it is assumed that the breakdown of EA to form products is slower than (i) formation of EA and (ii) breakdown of EA to re- form E and A. This means that the maximal reaction rate Vmax = k3[Et]. 3) Assumption 1: The concentration of the EA complex is constant (Steady State) 4) Assumption 2: Only the initial rate V0 is measured. This means that k4 can be ignored and there is no product inhibition or reaction reversibility. 5) Assumption 3: [A] >> [E] and the A concentration is considered constant. 12 6 points (a) Describe in details the difference between competitive, noncompetitive and uncompetitive inhibitors. (b) Give a shortly description on how the identity of an unknown inhibitor can be established through simple activity measurements. a) Competitive inhibitors work by binding at the active site on the enzyme. They compete with substrate for the active site and prevent the substrate from binding. Their structure is similar to that of the substrate since they are binding at the same site. The presence of these inhibitors causes an increase in Km of the enzyme, but leaves Vmax unaffected. Pure noncompetitive inhibitors are similar to uncompetitive inhibitors. Like uncompetitive inhibitors, they have a separate binding site on the enzyme and can bind the enzyme-substrate complex. However, they can also bind the enzyme when substrate is not bound. Presence of a noncompetitive inhibitor will decrease Vmax and will not affect Km. Uncompetitive inhibitors differ from competitive inhibitors in that they have a separate binding site on the enzyme. Also, they only bind to the enzyme when substrate is bound to the enzyme. Uncompetitive inhibitors decrease both Vmax and Km. b) In the first step the enzyme is assayed with variable substrate concentrations and a Lineweaver-Burk plot is created (1/V vs 1/[A]). In the second step, this process is repeated in the presence of a certain amount of inhibitor. A new line is added to the Lineweaver-Burk plot. This step can be repeated several times with different amounts of inhibitor. The different steps/experiments with generate a set of lines. For competitive inhibition all lines will dissect on the y-axis (1/Vmax). For noncompetitive inhibition the lines will dissect to, but now the point of dissection is in one of the quadrants (mixed noncompetitive) or on the x-axis (pure noncompetitive). For uncompetitive inhibition all lines are parallel and do not dissect. 13 3 points What does a large KI value for a competitive inhibitor indicate about the potency of the inhibition? Note that the equation that relates KI to values from the graphs is shown below.