Week 5: Protein Study Techniques and Enzyme Behavior Flashcards
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 1×107 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.
How can competitive and noncompetitive inhibition be distinguished in terms of KM?
The apparent value of KM increases with a competitive inhibitor, while it stays the same with an uncompetitive inhibitor.
Enzyme inhibition can be overcome under some, but not all, conditions. Under what conditions can inhibition be overcome?
A competitive inhibitor can always be overcome by the addition of enough substrate, while a noncompetitive inhibitor can never be overcome.
Because for a competitive inhibitor, the substrate and inhibitor compete for free enzyme molecules, adding enough substrate will allow all enzyme molecules to be in the ES form. This will allow the enzyme to run at its maximal velocity even in the presence of inhibitor.
In contrast, a noncompetitive inhibitor will bind to the ES complex. The presence of any inhibitor in this case will thus reduce the total amount of enzyme in the productive ES form, which will prevent the enzyme from running at its maximal velocity.
________ is the first step in protein purification. The various parts of cells can be separated by ______. This is a useful step because proteins tend to occur in given _______.
Disruption of cells
centrifugation
organelles
The purpose of cell rupture is to…
access the proteins within
First step -
release protein from cells and organelles
_________ used to break open cells
Homogenization
Differential Centrifugation
* Ruptured cells centrifuged ______
* As you ______ the force of gravity, _______ are in the pellet
* Differential spin speeds allow for ________
several times
increase, smaller and smaller components
particle separation
“Differential” means we are doing a _______ to get to the components we want.
series of steps
600 x g (600 times the force of gravity) - only the _______ will pellet out (be in the ______ at the bottom of the test tube). This would include _________ and _______
biggest pieces
precipitate
unbroken cells, cell nuclei
15000 x g - only _________ will pellet out, such as ______
smaller pieces
mitochondria
100000 x g - this will pellet out _______ such as _________
small items
ribosomes, membrane fragments
Often what we want is not in the _____ but is in the _______ (still ______ in the liquid). We can spin to the speed needed to ______ the things we don’t want and then work with the _______ that are still in the supernatant.
pellet, supernatant
dissolved
pellet out
soluble components
When would you choose to use a Potter–Elvehjem homogenizer instead of a blender?
If you needed to maintain the _______ of the subcellular organelles, a Potter–Elvejhem homogenizer would be better because it is _______.
Ex.
structural integrity
more gentle
The tissue, such as liver, must be soft enough to use with this device.
General order of protein purification:
- Crude Extract
- Salt precipitate
- Ion-exchange chromatography
- Molecular-sieve chromatography
- Immunoaffinity chromatography
Total activity is the total amount of _______ in a sample
target protein
Specific activity is…
It is a measure of _____
total activity / total amount of protein
protein purity
Salting Out (________)
* Purification based on _______ in salt
solutions
* Dependent on ________
salt fractionation
differential solubility
overall charge, ionic strength, and polarity of protein
In salting out, salt ions ________ allowing protein molecules to ________
attract water away from protein (have greater charge density)
interact with each other and aggregate
When enough salt is added in salting out protein ______ in _______ conformation
precipitates in native conformation (without denaturing)
In salting out, proteins with ________ precipitate first
greater hydrophobicity
In protein purification, _______ decreases with each step as _________. ________ increases with each step as ________
total protein, more contaminants are removed
specific activity, purity improves
A competitive inhibitor will alter the _______ of an enzyme-substrate combination but will not alter the _______. On a Lineweaver-Burk plot for a competitive inhibitor, the plots for separate experiments at different inhibitor concentrations will intersect on the ______.
The intersection will be at the point _____
apparent KM
apparent Vmax value
1/V axis
1/[S] = 0 and 1/V = 1/Vmax
A noncompetitive inhibitor will alter the ________ of an enzyme-substrate combination but will not alter the __________. On a Lineweaver-Burk plot, data collected at two different inhibitor concentrations will form plots that intersect on the _________
The intersection will be at the point _______.
apparent Vmax
apparent KM value
1/[S] axis
1/[S] = -1/KM and 1/V = 0
If you had a protein X, which is a soluble enzyme found inside the mitochocondrion, and you wished to separate it from a similar protein Y, which is an enzyme found embedded in the Golgi apparatus, what two techniques would permit you to complete this separation?
First technique to apply: Harsh homogenization of the cells (Harsh homogenization will liberate soluble cytosolic proteins and soluble organellar proteins into the cell buffer)
Second technique to apply: Centrifugation of 15,000 x g
(Spinning at 15,000 × g will pull down membranes (such as the Golgi apparatus containing protein Y) into the pellet. This will leave protein X in the supernatant, separated from protein Y)
Design an experiment to purify protein X on an cation-exchange column. Protein X is contained in a protein homogenate at pH 7. Protein X has an isoelectric point of 7.5.
1.
2.
3.
4.
- Decrease the pH of the protein solution to 6.0.
- Pour the protein homogenate at its proper pH through the column.
- Pour a salt gradient from low salt concentration to high salt concentration through the column.
- Collect the fraction containing protein X that comes out during the previous step.
Because the protein has an isoelectric point of 7.5, decreasing the pH to 6.0 will make the protein net positively charged. Pouring the pH 6.0 cell homogenate through an cation-exchange column will cause the positively charged proteins to stick to the column. Other materials will elute through the column. Increasing the salt concentration of the buffer that elutes through the column will disrupt ion-pair interactions, allowing the charged protein X to release from the column and elute out of the column. The fraction containing protein X will be subsequent to the start of the high salt wash.
Given that the isoelectric point of protein X is 7.5, the protein will have a net negative charge for pH > 7.5. Negatively charged proteins do not interact strongly with a cation-exchange column (which is itself negatively charged).
Starting at high salt concentration for the elution will cause all proteins bound to the cation exchange column to elute from the column. This procedure will not purify the column-bound proteins.
The first step in a western blot is the _______ via _________.
The next step takes the gel from the electrophoresis, and _______ from the gel onto a _______.
Once the proteins are transferred, they are _______, which binds to the target protein.
Then they are _________, which binds to the primary antibody, making it visible.
Lastly, the bands are made visible by reacting with the substrates for the _______, or they are visualized with a fluorometer or x-ray paper.
separation of proteins, electrophoresis
transfers the proteins, thin membrane of nitrocellulose or other absorbing compound
incubated with the primary antibody
incubated with the secondary antibody
secondary antibody enzyme tag
Other things being equal, what is a potential disadvantage of an enzyme having a very high affinity for its substrate?
The enzyme-substrate complex will be in a deep energy well, meaning that the enzyme-substrate complex will be ______.
High affinity of the enzyme for the substrate will _______ of the forward reaction.
more stable
increase the activation energy
High affinity of the enzyme for the substrate will decrease the energy of the enzyme-substrate complex, therefore, increasing the energy required to form the enzyme-transition state complex (the activation energy of the forward reaction).
The enzyme-substrate complex will be in a deep energy well, meaning that the enzyme-substrate complex will be more stable and, therefore, less reactive.
Tighter binding of the enzyme to its substrate does not increase the energy of the enzyme-transition state complex, although it does change the size of the activation energy to form this complex.
It is unlikely that distortion of the active site due to tight binding would occur. The enzyme is much larger than the substrate and thus not likely to deform.
For an enzymatically catalyzed reaction in which the measured values for KM and Vmax are, respectively, 13.9 mM and 1.02 mM min-1, what is the value of the turnover number? The concentration of enzyme used in the reaction is 50.0 μM.
20.4 min^-1
The turnover number for an enzyme is kcat = Vmax /[E] tot . Since the units of Vmax are mM min-1, it makes sense to convert [E] tot from μM to mM: [E] tot = 0.0500 mM.
The calculation of the turnover number is kcat = (1.02 mM min-1)/(0.0500 mM) = 20.4 min-1.
For an enzyme that obeys Michaelis-Menten kinetics, what is the reaction velocity, V, observed at the following substrate concentrations?
Express the result as a percentage of Vmax.
(a) If [S] = KM, then V/Vmax =
(b) If [S] = 10.0KM, then V/Vmax =
50%
90.9%
Substituting [S] = KM into this equation returns a value of 1/2. That is, the reaction’s velocity is 50% of the maximal velocity.
Substituting [S] = 10.0KM into this equation returns a value of 0.909. That is, the reaction’s velocity for this substrate concentration is 90.9 % of the maximal velocity.
For the hypothetical reaction,
3 A + 2 B –> 2 C + 3 D
the rate of the reaction was experimentally determined to be:
Rate = k[A]^2[B]^2
What is the order of the reaction with respect to A?
What is the order of the reaction with respect to B?
What is the overall order of the reaction?
2
2
4
Salting out is a process whereby a _______ is used to ______ of a protein until it comes out of solution and can be centrifuged. The salt forms ________ in the solution, which leaves less water available to hydrate the protein. ________ begin to interact between protein molecules, and they become insoluble.
highly ionic salt
reduce the solubility
ion–dipole bonds with the water
Nonpolar side chains
Their ______ and _______ make some proteins more soluble than others in ammonium sulfate. A protein with _______ on the surface is more soluble than ________ on the surface.
amino acid content
arrangements
more highly polar amino acids
one with more hydrophobic ones
To purify a protein, many techniques are used, and often _______ are used.
several different chromatography steps
Two phases of chromatography:
Stationary phase: selectively retards flow of
sample, effecting separation of components
Mobile phase: soluble (liquid) portion that migrates thru
stationary phase
In column chromatography, stationary phase is _______ and mobile phase is _______
packed in column
eluent (fluid used to elute a substance)
Size exclusion or gel-filtration chromatography is separation based on _____ and stationary phase is _________
size
cross-linked gel particles
In size exclusion chromatography, large molecules are ______ from gel pores, thus _______. Smaller molecules ______, thus _________
excluded from gel pores, migrate to bottom faster
partially or fully included in pores, migrate more slowly
In size-exclusion chromatography, the ______ elute first; the ____ elute last. ______ are excluded from the interior of the gel bead so they have _________ to travel. Essentially, they _______ and elute first.
largest proteins, smallest
Larger proteins, less available column space
travel a shorter distance
Sephadex has an exclusion limit of 80,000 molecular weight for globular proteins. If you tried to use this column material to separate alcohol dehydrogenase (MW 150,000) from -amylase (MW 200,000), what would happen?
Both proteins would elute in the void volume together and would not be separated.
Sephadex has an exclusion limit of 80,000 molecular weight for globular proteins, could you separate -amylase from bovine serum albumin (MW 66,000) using this column?
Yes, the -amylase would come out in the void volume, but the bovine serum albumin would be included in the column bead and would elute more slowly.
Affinity chromatography is a powerful column procedure based on _______. High purification often done _______
specific binding of molecules to ligand (best to pick a substance target substance would bind with)
done in one step
Ion-exchange chromatography is the method of separating substances on basis of ______.
charge
Cation exchanger: resin has ______
* Binds to ______ flowing through
net negative charge
(+) charged molecules
Anion exchanger: resin has ________
* Binds to _______ flowing through
net positive charge
(-) charged molecules
A cation exchanger binds _____. Hence the resin in the column must have a _______
cation
negative charge
In a cation exchange column, we need to find a pH where our target protein has a _________ but where our non-target proteins have ________ so that we achieve separation.
positive charge
neutral or negative charges
In ion exchange chromatography, protein then displaced by ______.
addition of high salt
Molecule carrying highest opposite charge will ______ to resin and thus ______.
bind strongest
elutes last
HPLC gives ______ purifications
gives fast and clean
Reverse phase HPLC: stationary phase _____ and mobile phase is ___
* ______ columns
* Take _____ for separation
nonpolar, polar
High-resolution
less time
What is the basis for the separation of proteins by the following techniques?
Gel-filtration chromatography
Affinity chromatography
Ion-exchange chromatography
Reverse phase HPLC
Size
Specific ligand-binding ability
Net charge
Polarity
Why do most people elute bound proteins from an ion-exchange column by raising the salt concentration instead of changing the pH?
A challenge is that extremes of pH can denature the proteins, so raising the salt concentration is often the way to go.
Design an experiment to purify protein X on an anion-exchange column. Protein X has an isoelectric point of 7.0.
Set up an anion-exchange column, such as Q-sepharose (quaternary amine). Run the column at pH 8.5, a pH at which the protein X has a net negative charge. Put a homogenate containing protein X on the column and wash with the starting buffer. Protein X will bind to the column. Then elute by running a salt gradient.
An amino acid mixture consisting of lysine, leucine, and glutamic acid 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. Which of these amino acids will be eluted from the column first? Will any other treatment be needed to elute one of these amino acids from the column?
Glutamic acid will be eluted first because the column pH is close to its pI. Leucine and lysine will be positively charged and will stick to the column. To elute leucine, raise the pH to around 6. To elute lysine, raise the pH to around 11.
An amino acid mixture consisting of phenylalanine, glycine, and glutamic acid is to be separated by HPLC. The stationary phase is aqueous and the mobile phase is a solvent less polar than water. Which of these amino acids will move the fastest? Which one will move the slowest?
A nonpolar mobile solvent will move the nonpolar amino acids fastest, so phenylalanine will be the first to elute, followed by glycine and then glutamic acid.
