BioChem Test 2 - SU 2014rename
evan406's version from 2016-03-23 18:00
|What methods are available to detect the folding or unfolding of a protein/enzyme?||1. The unfolding of an enzyme would coincide with a loss in activity. Therefore following the change in activity would be enough to study the unfolding processes. |
2. Absorption spectroscopy could be used if the protein/enzyme contains a colored cofactor that changes properties when the protein/enzyme unfolds
3. Circular dichroism can be used. This technique is sensitive to the type of secondary structure present in the folded protein: α-helix, β-sheet or random coil. Upon unfolding all the α-helices and β-sheets will be lost, converted to random coil, resulting in a large spectral change.
|Steps of GroEL-GroES chaperonin function in order, starting with an unfolded polypeptide and ending with a polypeptide in its native conformation?||1. A GroEL ring binds an unfolded polypeptide and 7 ATP |
2. GroES binds to GroEL, causing a conformational shift and displacing the polypeptide into the cavity
3. ATP is hydrolyzed and the polypeptide, isolated from interfering contacts, folds
4. The GroES “cap” dissociates from the GroEL ring
5. ADP and the folded polypeptide are released as ATP and GroES bind to the other GroEL ring
|What are the steps when Eukaryotic secretory proteins are synthesized and translocated via the endoplasmic reticulum?||1. signal sequence synthesis on ribosomes |
2. SRP binds signal sequence and subsequently binds SRP-receptor.
3. signal sequence removed
4. glycosylation in the ER lumen
5. ribosome dissociates
|Describe the role of ubiquitin in the breakdown of cellular proteins||-A chain of Ubiquitin molecules is covalently attached to the Lys residue of the target protein.|
- Three separate enzymes, E1, E2, and E3 are involved in this process.
- A polyubiquintylated protein is first recognized by the 19S regulatory particle and the ubiquitin is cleaved off.
- The protein is then fed through the base complex, is unfolded and digested into short peptides that are released into the cytosol where they are cleaved into individual amino acids by peptidases.
|For the binding of a ligand to a protein, what is the relationship between the Ka (association constant), the Kd (dissociation constant), and the affinity of the protein for the ligand?||Ka = 1/Kd|
The larger the Ka , the higher the affinity of the protein for the ligand.
|structure of heme c?|
|Explain how this heme c cofactor provides a capability that is not available to a protein made up from just regular amino acids. Why did nature select heme c and not a single Fe ion?||-Amino acids are not capable to reversibly bind oxygen. This capability is provided by the heme. |
-The electron donating character of the N-ligands prevents the Fe2+ from being oxidized when O2 binds. This is unique for heme c.
-Binding of oxygen to a single Fe2+ ion would result in the formation of reactive oxygen species that can damage biological structures.
|Although the heme c gives myoglobin and hemoglobin a new capacity, the protein environment also modifies the properties of the heme c. Explain how that is achieved.||-The protein environment changes the relative affinity of the heme for oxygen. |
-A histidine residue is able to form a hydrogen bond with the bound oxygen increasing the affinity for oxygen relative to that of free heme c.
|Define allosteric interaction.||The binding of a molecule to a binding site affects binding properties of another site on the protein.|
effect of oxygen pressure on binding to hemoglobin
|The blood pH drops from 7.4 to 7.2.||from B to C|
|The blood CO2 concentration increases.||from B to C|
|The concentration of 2,3-bisphosphoglycerate (BPG) increases during acclimation to high altitude.||from B to C|
|An infant's fetal hemoglobin (HbF) is replaced by normal adult hemoglobin (HbA).||from B to C|
|Hemoglobin is isolated from red cells and stripped of 2,3-bisphosphoglycerate (BPG).||from B to A|
|Tetrameric hemoglobin is dissociated into its subunits.||from B to A|
|Explain why the binding affinity of hemoglobin oxygen changes when the pH changes||-Due to the Bohr effect there is a change in affinity for oxygen when the pH changes. |
-low pH in tissues due to increased CO2 concentration and the resulting formation of bicarbonate CO2+H2O -> HCO-3+H+,
-certain amino acids become protonated an can form ion bridges with other amino acids. These bridges are only formed in the T-state.
-equilibrium between the R- and T-states shift towards the T-state and hemoglobin binds oxygen less tightly releasing most of the oxygen bound.
-The formation of a carbamate at the N-terminus due to reaction with CO2 has a similar effect also stabilizing the T-state.
-In the lungs the pH goes up and the opposite effect is detected where the protein equilibrium shifts now more towards the R- state and will bind much more oxygen than possible at the lower pH.
|Why are enzymes such good catalysts?||-Formation of ES complex lowers the Ea by stabilizing the transition state or providing an alternative reaction pathway with lower Ea. |
-This makes the enzyme highly specific for its substrate but also for the reaction it will perform.
-It does this all under mild conditions of pH, temperature and pressure in aqueous solution.
|What are the benefits of measuring the initial rate of a reaction V0?||-At t=0 no product will have been formed and its reverse reaction is negligible. |
-changes in [S] are negligible, so [S] can be treated as a constant.
|What does the steady state assumption, as applied to enzyme kinetics, imply?||The enzyme-substrate complex is formed and broken down at equivalent rates.|
|What is the turn over number? How do you calculate it?||The turn over number is the number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation. |
|For a reaction that can take place with or without catalysis by an enzyme, what would be the effect of the enzyme on the: |
A. standard free energy change of the reaction?
B. activation energy of the reaction?
C. initial velocity of the reaction?
D. equilibrium constant of the reaction?
|A. No change|
D. No change
|You measure the initial rate of an enzyme reaction as a function of substrate concentration in the presence and absence of an inhibitor. The following data are obtained:|
a) What is the Vmax in the absence of inhibitor?
b) What is the Km in the absence of inhibitor?
c) When [S] = 0.0004, what will V0 be in the absence of inhibitor? (Use the M&Mequation to calculate this value)
d) When [S] =0.0004, what will V0 be in the presence of inhibitor?
e) What kind of inhibitor is it likely to be?
|a) The measured V0 appears to level out at 100. Therefore Vmax = 100|
b) 1⁄2 Vmax = 50. This rate is measured at [S] = 0.0002 = Km
d) apparent Km with inhibitor: 0.0005
e) The original Vmax is still obtained at high [S]: competitive
Lineweaver-Burk plots (reversible inhibition types)
|What type of inhibitor and how is Vmax and Km affected?||Competetive. Competition for the active site, Vmax not changed|
|What type of inhibitor and how is Vmax and Km affected?||Uncompetetive. Both Vmax and Km changed with same amount (α’|
|What type of inhibitor and how is Vmax and Km affected?||Mixed noncompetitive. Both Vmax and Km changed with different amounts|
|What type of inhibitor and how is Vmax and Km affected?||Pure noncompetitive. Km not changed|