HYPS- Biochem Flashcards
a helices characteristics
rodlike structures in which the peptide chain coils clockwise about a central axis. The helix is stabilized by intramolecular hydrogen bonding that occurs between carboxyl oxygens and amino hydrogens located four residues away from each other. Typically amino acid side chains point away from the helix’s core, allowing for interaction with the cellular environment. There is extension of hydrophilic side chains away from the core of the protein toward the aq environment of the cell.
Beta pleated sheets
Peptide chains form rows that are held together by initramolecular hydrogen bonding that occurs between the carboxyl oxygen on one peptide chain and the amino hydrogen on another. Pleating maximizes H bonding in the structure. B pleated sheets may have 1 side of a pleated sheet w/ hydrophobic side chains while the other may be hydrophilic.
Why are alpha helices more easily degraded than beta sheets
alpha helix has peptide bonds more accessible to peptidases , enzymes that cleave peptide bonds. Beta pleated sheets are more stable and expose far fewer residues to the cellular environment
Mechanism of CJD
prions convert more soluble degraded alpha helices to less soluble more difficult to degrade beta pleated sheet conformation. This causes build up and formation of plaques, which can lead to loss of cell function and ultimately cell death.
A point mutation changes a cysteine residue to an alanine residue. How might this affect protein structure?
cysteine residues contain sulfhydryl groups that form disulfide bonds, a type of tertiary structure. disulfide bonds can also stabilize local secondary and other tertiary structures. they also help forma hydrophobic core within proteins, facilitating hydrophobic interactions. A change in a single cysteine residue may interrupt a key disulfide bridge, resulting in the inability of a protein to maintain its secondary and tertiary structures or its hydrophobic core. Ultimately this may also disrupt quat structure by changing interactions between multiple peptides. Thus, this single change can cause disruption of protein structure and function.
A protein is treated with 6 M of HCl. what levels of protein structure are most likely disrupted by treatment with this solution?
Results in denaturation of a protein, or loss of secondary tertiary and quaternary structure. Concentrated strong acid may also permit hydrolysis of the peptide bond, interrupting primary structure.
A point mutation causes a single leucine residue to be substituted for an isoleucine residue in the transmembrane section of a G protein coupled receptor. How might this change affect overall protein structure and function?
A G protein coupled receptor contains a transmembrane domain that consists mainly of hydrophobic residues. Because this mutation is not in the binding site of the receptor, it is unlikely to affect substrate binding. The specific mutation described subs one hydrophobic aa for another, which should cause minimal changes in the secondary and tert structures of the protein. Therefore, this mutation is unlikely to affect the structure or function of the peptide.
A lineweaver Burk plot of HMG-CoA reductase, the rate limiting enzyme of cholesterol synthesis is competitively inhibited by Statins. How would the use of a statin affect the Lineweaver burk plot?
Competitive inhibitors compete with the substrate for binding to the active site of an enzyme. The presence of these inhibs results in fewer active sites being available to act on the substrate, resulting in decreased enzyme activity. Comp inhib can be overcome by increasing the substrate concentration and can potentially outcompete the inhibitor, allowing the enzyme to regain vmax. Therefore, while it may take more substrate to reach vmax, the value of vmax itself DOES NOT CHANGE.
Increasing the substrate conc allows the substrate to outcompete the inhibitor and the enzyme can regain the same maximal velocity. However, because a higher conc of substrate is necessary to reach the same vmax, the Km of the enzyme will also INCREASE. Km can be thought of as a measure of the affinity between an enzyme and its sub. a HIGHER KM indicates DECREASED affinity between an enzyme and its substrate. SO. graph would have no change in y intercept, but Km increases (so point on x axis moves closer to origin)!!!
How would the Lineweaver-Burk plot change in the presence of a noncompetitive inhibitor?
A noncompetitive inhibitor binds to an allosteric site, rather than the active site. This changes the conformation of the active site, leading to a decrease in the efficiency of enzyme catalysis, which results in a decrease in vmax. However, the Km WILL NOT change due to noncompetitive inhibition because any copies of the enzyme still in the active conformation can bind the substrate with the same affinity. Therefore, compared to the line without inhibitor, the line with a noncompetitive inhibitor will have the same x-intercept, but a higher y intercept.
How would the Lineweaver Burk plot change in the presence of an uncompetitive inhibitor
An uncompetitive inhibitor binds to an allosteric site, but only when the substrate is already bound to the enzyme. This results in an increased affinity for substrate bound to enzyme and decreased dissociation of the enzyme substrate complex. Therefore, uncompetitive inhibition results in a decrease in both VMAX AND KM. Compared to the line without an inhibitor , the line with an uncompetitive inhibitor will have a more negative x intercept (increased affinity, lower km) and a higher y intercept (decreased vmax). so it will shift upwards and to the left.
In a Michaelis menten plot of HMG-CoA reductase, the rate limiting enzyme of cholesterol synthesis is competitively inhibited by Statins. How would the use of a statin affect the MM plot?
MM plots have same info as LB plots but have diff axes. The y axis in a MM plot is the rate or velocity, v, and the x axis is the substrate conc, [s]. Monomeric enzymes will result in a hyperbolic shape in the graph, reaching a plateau at vmax. A comp inhibitor will result in a lowered plateau; however, the curve will not shift or be stretched left or right because the Km does not change.
