Week 5: Protein Study Techniques and Enzyme Behavior Flashcards
(198 cards)
The term “salting out” refers to a phenomenon whereby a highly soluble salt is added to a protein solution to _______ the solubility of the protein. This change in protein solubility can be exploited by the use of __________ to concentrate and purify the protein.
decrease
centrifugation
Salting out works because increasing the salt concentration permits the _________ of individual protein molecules to interact with each other, allowing ________
hydrophobic portions
aggregation and subsequent precipitation.
Which scenario is the best one to partially isolate a soluble cytosolic protein using only one differential centrifugation spin?
Break open the cells and centrifuge at 100,000 × g. The protein will be in the supernatant.
After disrupting the cell membrane, soluble cytosolic proteins will be free of the cells. Spinning at 100,000 × g will pull down the cell microsomal fraction and debris, leaving soluble cytosolic proteins in the supernatant. Further purification will be needed to separate the target protein from other soluble cytosolic proteins.
_________ is the basis for the separation of proteins by the reverse phase HPLC technique.
Degree of hydrophobicity
Reverse phase HPLC is a form of high performance liquid chromatography, in which the stationary phase is ______ and the mobile phase is a _______. Proteins with a higher degree of _______ will interact to a greater degree with the _______ and take _______ to travel through the column.
nonpolar, polar liquid
hydrophobicity
nonpolar stationary phase
longer
_______ are not useful bases for separating proteins.
Protein primary and secondary structures
The first proteins to elute on a gel filtration column are the ______ in the sample.
largest proteins
Larger proteins have a _______ through the column as compared to smaller proteins.
shorter path length
Smaller molecules enter the pores of the gel causing them to move more slowly, while large molecules cannot enter the pores of the gel because they are too big. Thus, large molecules proceed through the column at a faster rate due to their inability to be absorbed by the gel. This effectively shortens the distance that the larger proteins have to travel through the column relative to the smaller proteins.
A compound can be eluted from an ion-exchange column by _______ or by _________.
raising the salt concentration
changing the pH
When the salt concentration is increased the bound proteins are _______. The advantage of using salt is that ______. The disadvantage is that _______.
outcompeted by the salt ions
it is cheap
it might not be specific for a particular protein.
Sephadex G-75 has an exclusion limit of 80,000 molecular weight for globular proteins. If you tried to separate trypsinogen (MW 20,000) from β-amylase (MW 200,000) using this resin in a column, what would happen?
The 80,000 MW cutoff of Sephadex G-75 will exclude proteins of molecular weight greater than 80,000 from the bead matrix. Thus, β-amylase will not enter the bead matrix and elute first in the void volume. Trypsinogen (MW 20,000) is able to enter the bead matrix, hence will elute second.
An amino acid mixture consisting of lysine, glutamic acid, and leucine is to be separated by ion-exchange chromatography, using a cation-exchange resin at pH 3.5, with the eluting buffer at the same pH. The first amino acid to be eluted will be ________.
The other two amino acids will remain on the column because they are _________.
In order to elute these two proteins from the column it will be necessary to
glutamic acid
positively charged
raise the pH first to 7 to elute leucine, and then raise it to 11 to elute lysine.
An amino acid mixture consisting of alanine, isoleucine, and aspartic acid is to be separated by normal phase HPLC. The stationary phase is polar and the mobile phase is a nonpolar solvent. The first amino acid to be eluted will be ________.
The last amino acid to be eluted will be _________.
isoleucine
aspartic acid
A nonpolar mobile phase will result in nonpolar amino acids travelling more rapidly through the column than the more polar amino acids.
A nonpolar mobile phase will result in nonpolar amino acids travelling _______ through the column than the more polar amino acids.
more rapidly
The speed of migration of a protein undergoing electrophoresis is determined by:
Tertiary structure
Size
Shape
Net charge
The denaturation of proteins by SDS leads to all the proteins having roughly the same shape, which is a _________.
SDS breaks ______. Therefore, multisubunit proteins can be analyzed as the _________.
The negative charge on SDS imparts a _______ to the SDS-protein complex.
Proteins are separated based on ______.
random coil
all noncovalent interactions
component polypeptide chains
net negative charge (SDS binding gives proteins a net negative charge)
size alone
Biological catalysts are orders of magnitude ________ as catalysts.
Nonenzymatic catalysts are __________, whereas enzymes are _______.
All enzymes are…
more effective
low molecular weight compounds
high molecular weight proteins
All enzymes are catalysts, but not all catalysts are enzymes.
Catalase breaks down hydrogen peroxide about 1×10^7 times faster than the uncatalyzed reaction. If the latter required one year to achieve a certain degree of completion, how much time would be needed by the catalase-catalyzed reaction?
3.15 sec
tcat = (1 year) × (365 days/year) × (24 hr/day) × (60 min/hr) × (60 sec/min) / 1×107 = 3.15 sec.
The reaction of glucose with oxygen to produce carbon dioxide and water,
Glucose + 6O2 –> 6CO2 + 6H2O
has a ΔG° of -2880 kJ mol-1, making it a strongly exergonic reaction. However, a sample of glucose can be maintained indefinitely in an oxygen-containing atmosphere. How can these two facts be reconciled?
Although the reaction is exergonic, the activation energy barrier for the reaction is so high that the reaction occurs very slowly.
In order for the reaction to proceed, it must be possible for the reactants to overcome the reaction’s activation energy. If the activation energy barrier is very high, the forward rate constant for the reaction will be very slow.
The presence of a catalyst has _______ on the standard free-energy change. The presence of a catalyst affects the ______, which is a ______ property. The standard free-energy change is a _______ property that ___________ on the reaction rate.
no effect
rate of a reaction
kinetic
thermodynamic
does not depend
The rate of the reaction ATP → ADP + phosphate will be enhanced by _______ as the rate for ADP + phosphate → ATP.
the same factor
The rate constant k for a reaction depends on the activation energy EA according to k = Ae-EA. Because the forward and reverse reactions must cross the same peak energy point in the reaction path, EA,f and EA,r are both equally affected by the catalyst. Thus, the forward and reverse rate constants will both increase by the same factor in the catalyzed reaction.
In the lock-and-key model, the active site ________.
In the induced-fit model, the active site _________.
has well-defined shape
takes shape around substrate.
How is the KM related to substrate concentration when V = Vmax/2?
KM is equivalent to the substrate concentration.
KM is the Michaelis constant. When the reaction velocity, V, is half the maximum, Vmax, KM is equivalent to the substrate concentration. One interpretation of the Michaelis constant is that it is the substrate concentration at which half of the enzyme active sites are occupied by substrate.
How can competitive and noncompetitive inhibition be distinguished in terms of Vmax?
The Vmax remains unchanged with a competitive inhibitor, while it decreases with a noncompetitive inhibitor.
A competitive inhibitor competes with the substrate for the active site of the enzyme, therefore it can be overcome by high substrate concentrations. Hence, Vmax, the velocity when the enzyme is saturated with substrate is unchanged.
A noncompetitive inhibitor does not compete directly with the substrate as it binds to the enzyme at a site away from the active site. Therefore, increasing the substrate concentration does not overcome noncompetitive inhibition and Vmax decreases.