Unit Two Flashcards
1
Q
true or false: The tertiary structures of myglobin, α-hemoglobin and β-hemoglobin are nearly identical, therefore it is expected they will have nearly identical primary structure.
A
false
2
Q
In the ____ form of Hb, the iron atom is out of the plane of the porphyrin ring.
A
T
3
Q
The Bohr effect describes a lowered O2-affinity of Hb when the pH _____.
A
decreases
4
Q
Which of the following would be an example of a conservative amino acid substitution?
Arg replaced with Lys
Asp replaced with His
Gly replaced with Ser
Glu replaced with Gln
A
Arg replaced with Lys
5
Q
Which of the following statements about the T to R transition of Hb is FALSE?
A) The change in tertiary conformation results in a change in quaternary structure
B) His F8 residue moves between two Thr residues when O2 binds
C) Binding of O2 induces formation of a 6th coordination with the heme iron.
D) Binding of O2 pulls the iron atom into the plane of the porphyrin ring.
A
His F8 residue moves between two Thr residues when O2 binds
6
Q
During the T to R transition of Hb, Fe2+ drags the F-helix due to an electrostatic interaction with which amino acid residue?
A
His F8
7
Q
Which of the following statements regarding the effect of 2,3-BPG on Hb is FALSE?
A) BPG does not bind to Hb in the R state
B) BPG stabilizes Hb in the T state.
C) Fetal Hb binds O2 with lower affinity than that of adult (maternal) Hb.
D) In the absence of BPG the affinity of Hb for O2 in tissues is too high.
A
Fetal Hb binds O2 with lower affinity than that of adult (maternal) Hb.
8
Q
Highly active muscle generates lactic acid by respiration so rapidly that blood passing through the muscle drops in pH from 7.4 to 7.2. Would Hb bind O2 with higher or lower affinity at 7.2 than it does at pH 7.4?
A
lower
9
Q
An allosteric interaction between a ligand and a protein is one in which
A
binding of a molecule to a binding site affects binding properties of another site on the protein
10
Q
When oxygen binds to a heme-containing protein, the two open coordination bonds of Fe2+ are occupied by
A
one O2 molecule and one amino acid atom
11
Q
In the binding of oxygen to myoglobin, the relationship between the concentration of oxygen and the fraction of binding sites occupied can best be described as
A
hyperboplic
12
Q
Which statement about protein-ligand binding is CORRECT
A) The larger the Ka (association constant), the weaker the affinity
B) The larger the Ka, the faster is the binding
C) The Ka is equal to the concentration of ligand when all of the binding sites are occupied
D) The larger the Ka, the smaller the Kd (dissociation constant)
E) The Ka is independent of such conditions as salt concentration and pH
A
The larger the Ka, the smaller the Kd (dissociation constant)
13
Q
Myoglobin and the subunits of hemoglobin have
A) very similar primary structures, but different tertiary structures
B) no obvious structural relationship
C) very similar primary and tertiary structures
D) very different primary and tertiary structures
E) very similar tertiary structures, but different primary structures
A
very similar tertiary structures, but different primary structures
14
Q
In hemoglobin, the transition from T state to R state (low to high affinity) is triggered by
A
oxygen binding
15
Q
Which statement is NOT correct concerning 2,3-bisphosphoglycerate (BPG)?
A) It binds at a distance from the heme groups of hemoglobin
B) It is an allosteric modulator
C) It increases the affinity of hemoglobin for oxygen
D) It is normally found associated with the hemoglobin extracted from red blood cells
E) It is an allosteric negative modulator
A
It increases the affinity of hemoglobin for oxygen
16
Q
Carbon monoxide (CO) is toxic to humans because it
A
binds to the Fe in hemoglobin and prevents the binding of O2
17
Q
The coordinated nitrogen atoms in the heme prosthetic group have _____ character, which prevents conversion of the heme iron from the _____ state to the _____ state
A
electron-donating; Fe2+; Fe3+
18
Q
The change in hemoglobin-binding affinity for oxygen due to changing pH and CO2 concentration in the blood is known as the
A
Bohr effect
19
Q
Use the following equation to calculate the fractional saturation of myoglobin as it moves from the cell surface where pO2 is 10 torr to the mitochondria where pO2 is 1 torr.
YO2 = pO2/ (K + pO2)
Based on the fractional saturation change, what can you conclude about myoglobin?
A) Myoglobin’s affinity for oxygen (Ka) changes as ligand concentration changes
B) Myoglobin remains saturate with oxygen under this conditions
C) Myoglobin releases oxygen as the partial pressure decreases aiding in diffusion
D) Myoglobin’s binding saturation increases under this conditions
A
Myoglobin releases oxygen as the partial pressure decreases aiding in diffusion
20
Q
How well would myoglobin facilitate the diffusion of oxygen inside the cell (pO2 at membrane 10torr, pO2 at the mitochondria 1 torr) if it’s p50 was 50torr?
A) Not well, it won’t bind much oxygen at the cell surface and therefore will slow down diffusion to the mitochondria
B) Very well, it will saturate with oxygen in the cell surface and facilitate diffusion to the mitochondria and deliver about 50% of the oxygen it caries in the
C) Not well, it will saturate with oxygen in the cell surface and not facilitate diffusion to the mitochondria
D) None of these choices
A
Not well, it won’t bind much oxygen at the cell surface and therefore will slow down diffusion to the mitochondria
21
Q
Which alpha helixes contain the two functional amino acids for myoglobin?
A
F and E
22
Q
Which histidine helps coordinate the iron in the protoporfirine prosthetic group
A
proximal
23
Q
The major secondary structure found in myoglobin is
A
alpha helix
24
Q
Which amino acid side chain increases binding affinity to O2 with respect to other molecules like CO, NO, and H2S ?
A
distal
25
Oxygenated globin has color while deoxygenated globin has a color
red, blue
26
Which histidine directly interacts with oxygen and prevents iron oxidation to Fe3+
distal
27
how does heme interact with iron
She said she wants to know how the heme is coordinated by iron: all hemes contain a central iron ion, which is coordinated by the four pyrrole nitrogen atoms.
metal iron atom in heme is what binds directly to oxygen. The iron atom lies at the center of the protoporphyrin (organic component of heme group) and is bound to 4 nitrogen atoms.
28
how does the globin interact with iron
Globin prevents the oxidation of iron to Fe3+ due to steric hinderance
29
what is the function of heme in oxygen binding
The heme group binds oxygen by bending, which allows the reversible binding of oxygen (maybe on that second part)
In the deoxy state iron is 2+ . In this state Fe is too large to fit in the plane with protoporphyrin, so the Fe atom is found slightly below the plane (bent). Once Fe is bound with oxygen the size of Fe decreases so it can fit at the center of the protoporphyrin plane (straight line oxy state).
protoporphyrin is an organic ring system with iron at the center
30
what is the function of globin in oxygen binding
- oxygen transport and storage of oxygen
- globins require a prosthetic group to bind to oxygen (heme)
- globin prevents the oxidation of iron to Fe3+ by introducing steric hindrance on one side of the heme plane so interaction can be prevented (it protects one of the heme groups)
31
what is the function of proximal histidine and what number is it
His F8 is the proximal histidine
it prevents oxidation by blocking the binding site to Fe (binds directly to the iron of heme)
32
what is the function of distal histidine and what number is it
His E7 is the distal histidine
- improves the affinity for oxygen and decreases the affinity for other gases (like CO)
- does not directly interact with the HEME group but helps stabilize the binding of oxygen to the iron
- stabilizes oxygen AFTER binding to the iron
33
what is the importance of the Asp--His electrostatic interaction
this ionic interaction stabalizes the T-state. When O2 binds it breaks these interactions making it easier to change the structure from T-state to R-state.
- Asp and His act as pH sensors (why is this important, specifically the bohr effect ---> the pH difference between the lungs and metabolic tissues increases the efficiency of the O2 transport)
34
what is the protein structure (native fold) of myoglobin, and how does this explain the lack of cooperativity
- single polypeptide chain, folded into 8 alpha helixes
- monomer
- highest level of structure is tertiary
- one prosthetic group: heme
- myoglobin only has one heme group, thus one binding site. Because it is a monomer, it can not display cooperativity because there are no binding sites that can be altered after binding at one sight has occurred.
35
what does the binding curve of myoglobin tell you, and what would affect it
- binding curve is hyperbolic which tells you it is not cooperative
- from the binding curve you can determine the PO2 and the P50. P50 is the the partial pressure of oxygen required to to reach 50% saturation.
- The higher the P50 the lower the affinity for oxygen and vice versa
36
how do you find p50 for myoglobin
P50 is when Y=0.5
37
how do you find the fractional saturation for myoglobin
Y=PO2/(P50+PO2)
38
how does ligand concentration effect the curve or equation for myoglobin
myoglobin is more sensitive to binding when ligand concentration is low. As ligand concentration increases, it is less crucial to bind ligand and therefore sensitivity decreases, this is why myoglobin functions better as an oxygen-storage protein.
39
what does the binding curve for hemoglobin tell you
- the binding curve for hemoglobin is a sigmoidal binding curve, which tells you that it participates in cooperativity
- if the curve shifts left, there is an increased affinity for O2. Things that would cause a shift left include: decrease in pCO2, decrease in [H+], and decrease in temp
- if the curve shifts right, Hb has a decreased affinity for O2. Things that would cause a shift right include: increased pCO2, increase [H+] (which is an increase in pH), and increase in temp
40
how do you find the p50 for hemoglobin
Y=(pO2)^n/((pO2)^n+(P50)^n)
p50 is when it is 50 % saturated
41
how do you find fractional saturation for hemoglobin
Y=(pO2)^n/((pO2)^n+(P50)^n)
42
how do you find ligand concentration using the curve or fractional saturation for hemoglobin
Y=(pO2)^n/((pO2)^n+(P50)^n)
43
what is the protein structure (native fold) for hemoglobin and how does this explain its cooperativity
- Hemoglobin is a tetramer of two subunits (alpha 2 and beta 2) (the alpha and beta names have nothing to do with structure)
- each subunit is similar to myoglobin
- 4 polypeptide chains, 8 alpha helixes except for D in alpha Hb (secondary level of structure for each subunit)
- 4 prosthetic groups: 4 heme
- the main function of hemoglobin is to transport oxygen. Binding at one site leads to a conformational change at the other binding sites, increasing the affinity of oxygen
44
how does cooperativity take place in hemoglobin (molecular details) how is oxygen a positive modulator
1. Hemoglobin begins in the T state, this is the deoxygenated state (Fe 2+ is not aligned with the heme plane, causing a bend)
2. Oxygen binds to one of the subunits -- this drives the conformational change, BUT the proximal histidine moving is what actually forces the change
3. Hemoglobin subunit is now in the R state. In the R state Fe2+ is lined up in a straight line with the plane of heme. The R state has a high affinity for oxygen
4. The transition from T state to R state triggers changes in the other subunit, increasing their affinity for oxygen
this makes oxygen a positive modulator
45
what would happen to hemoglobin affinity and protein structure if there were changes in H+
The association of protons (H+ ions) with the amino acids in hemoglobin that cause a conformational change in protein folding, ultimately reducing the affinity of the binding sites for oxygen molecules.
- quaternary structure of hemoglobin is essential to its cooperativity. With no quaternary structure hemoglobin would be stuck in the high affinity state, because the T-state is only retained when all 4 subunits are having intramolecular interactions
46
what would happen to hemoglobin affinity and protein structure if there were changes in CO2
Increasing pCO2 decreases the pH, which would reduce the affinity for oxygen (curve shifts right)
47
what would happen to hemoglobin affinity and protein structure if there were changes in 2,3-BPG
decrease in 2,3-BPG results in higher affinity for oxygen (curve shifts left)
Increase in 2,3-BPG results in lower affinity for oxygen (curve shifts right)
48
what does the value of the hills coefficient represent
it indicates the degree of cooperativity between subunits. The hill constant will never be equal to the number of binding sites on a protein, because that would indicate infinite cooperativity.
hill constant=1 non-cooperative
hill constant >1 positive cooperativity
hill constant < 1 negative cooperativity (very rare)
49
what does the hill coefficient plot tell you about binding behavior (correlate to models concerted vs. sequential)
concerted model:
-for all 4 subunits, no intermediate states
- all subunits change conformation simultaneously
- characterized by an equilibrium between T and R state
- Disadvantage: not all proteins have identical subunits
- related to lock and key model
sequential model:
- many possible conformations of T and R states in one Hb protein
- each binding event causes conformational changes that increase affinity gradually
- related to the induced fit model
50
how are enzymes classified
enzymes are classified by the reactions they catalyze. They are divided into 7 classes
51
what are the different enzyme classifications
1. oxireductase
2. transferases
3. hydrolases
4. lyases
5. isomerases
6. isomerases
7. translocases
52
what are the thermodynamic parameters used to describe an enzyme catalyzed reaction
the reaction equilibria are linked to the standard free-energy change for the reaction
reaction rates are linked to the activation energy (delta G transition)
Keq=1; delta G= zero
Keq>1; delta G=negative
Keq<1; delta G= positive
first and second order reactions
I don't think this what she is actually looking for in this question
53
how is activation energy decreased by an enzyme
enzymes decrease the activation energy by stabilizing the transition state between the enzyme and the product.
enzymes do not effect the equilibrium point
the sum of the unfavorable activation energy (delta G transition) and the favorable binding energy (delta GB) results in lower net activation energy.
covalent interactions between the enzyme and substrate lower the activation energy
54
what is the difference between the lock and key model and the induced fit model
The lock and key model states that the active site of an enzyme precisely fits a specific substrate. The induced fit model states that the active site of an enzyme will undergo a conformational change when binding a substrate, to improve the fit.
55
how do enzymes increase the rate of reaction
they increase the rate of reaction by decreasing the activation energy
56
what can enzymes not change in a reaction
enzymes DO NOT change the equilibrium point or the [P] or [S]
57
what are the 5 ways enzymes increase the speed of a reaction
COME ADD MORE TO THIS
Binding substrate=proximity and orientation,
stabilization of
transition state
general A/B catalysis covalent catalysis
metal ion catalysis
58
what is the effect of pH on enzyme catalysis
enzymes have an optimal pH. a graph of enzyme activity and pH has a bell curve. At the top of the curve it is at optimal pH. Extremely high or low pH values outside the optimal range will denature the enzyme resulting in loss of function.
59
what is the effect of temperature on enzyme catalysis
enzymes work best at an optimal temperature. Increasing temperature will increase activity until a peak is reached, after that increases in temperature will denature the enzyme (bonds break)
as temperature decreases molecules move slower so the reaction will proceed more slowly, or will not function properly
60
what are the 4 equations you need for Michaelis-Menten calculations
Michaelis-Menten equation
Double reciprocal equation
Vmax= Kcat[Et] for catalytic constant
catalytic efficiency= Kcat/Km
sorry I am not typing these out rn but this is what shes looking for
61
for Michaelis-Menten what is the assumption of steady and why is it important
the steady state assumption is important because steady state assumes that the concentration of the ES is constant, at equilibrium, and the rate of ES formation = the rate of ES breakdown. The purpose of MM kinetics is to find the velocity/rate of the reaction at different [S]. To be able to calculate for velocity at different substrates we need to assume that ES concentration is constant and it is forming and breaking down at an equal rate
62
in Michaelis-Menten why is enzyme concentration kept constant during kinetic studies
changing the concentration of the enzyme would change the Vmax value. (more enzyme = more available active sites = larger Vmax, and vice versa). So, to make an MM plot, we keep enzyme concentration constant, with the only changing variable being [S], so that the Vmax value stays the same
63
In Michaelis-Menten why is velocity only measured during steady state conditions and early in a reaction
because later in the reaction the velocities will all be the same, giving a straight vertical line for the MM plot
I think im not really sure and i cant find it on google???
64
In Michaelis-Menten what is the meaning of Vmax, Km, Kcat, and what is constant
Vmax is max product rate when all E are in ES form.
Km = substrate req for 50% saturation of E. sometimes called affinity. High km means lots of [S] needed (low affinity) (substrate constant at Vmax)
Kcat - product formation High is fast low is slow
Km is a constant
65
what are the 4 different types of inhibitors
- competitive inhibition
- uncompetitive inhibition
- mixed inhibition
- noncompetitive inhibition
66
how do you calculate parameters using MM equation and the double-reciprocal equation
double reciprocal= lineweaver Burke plots.
X intercept = -1/km'
Y intercept= 1/Vmax
Slope = Km/Vmax
67
what is the mechanism for chymotrypsin. What happens in each step
Chymotrypsin is a digestive proteolytic enzyme that helps digest proteins.
Step 1: The polypeptide substrate binds non covalently with side chains of hydrophobic pocket (this correctly positions the peptide bond for attack)
Step 2: interaction between Ser and His generates a strong nucleophilic attack on Ser. The substrate forms a tetrahedral intermediate with the enzyme
Step 3: negative charge on the substrate carbonyl creates a highly unstable intermediate leading to a collapse of the tetrahedral intermediate. A double bond with carbon is reformed, which breaks the peptide bond. Amino leaving group is protonated by His, facilitating its displacement
Step 4: incoming water molecule is deprotonated by general base catalysis, creating a strongly nucleophilic hydroxide ion. Hydroxide ion attacks ester linkage of acyl-enzyme, generating a second tetrahedral intermediate
Step 5: the water molecule transfers its proton to His and its -OH to the remaining substrate fragment,
Step 6: the second peptide fragment is released: the acyl bond is cleaved, the proton is transferred from His back to Ser, and then the enzyme returns to its initial state
GENERAL SUMMARY:
- operates through a ping pong mechanism called covalent hydrolysis
- enzyme first forms covalent bond with target substrate, displacing the more stable part into the solution
- enzyme substrate complex is the enzyme intermediate
- intermediate interacts with water which displaces the remaining part of the initial substrate and reforms the initial enzyme
68
what is a catalytic triad
The main driving force for chymotrypsin is the catalytic triad. There are 3 amino acids residues in active site in a hydrogen bonded network: Ser 195, His 57, Asp 102. The catalytic action converts OH group of Ser 195 into a potent nucleophile. His acts as base catalyst to activate Ser and Asp stabilizes protonated Ser
69
what is an oxyanion hole
a pocket in the active site of an enzyme that stabilizes transition state negative charge on a deprotonated oxygen or alkoxide. Stabilizing the transition state lowers the activation energy necessary for the reaction (promotes catalysis).
In chymotrypsin the amide hydrogens of Ser195 and Gly193 form the oxyanion hole
69
In chymotrypsin, which amino acids are responsible for general acid catalysis, covalent catalysis, and when is each amino acid doing each
Ser 195: Covalent Catalysis- it attacks the C=O of the peptide bond, forms an oxyanion tetrahedral unstable transition state
His 57: General acid base catalysis: His first acts as a general base that deprotonates Ser 195 (a very weak acid) and makes serine a good nucleophile to attack C=O in the peptide bond
Asp 102: Electrostatic stabilization. Stabilizes His 57's conjugate acid and makes it a better base, allowing it to deprotonate the super weak acid, Ser 195, easier. Also stabilizes entire mechanism in general
70
in chymotrypsin what is the substrate binding effect
when the substrate binds, the side chain of the residues adjacent to the peptide bond to be cleaved nestles in a hydrophobic pocket on the enzyme, positioning the peptide bond for attack
71
in chymotrypsin what is the function of water
The intermediate reacts with water, which displaces the remaining part of the initial substrate and reforms the initial enzyme.
72
what does oxireductase do
transfer of electrons (hydride ions or H atoms)
involved in redox reactions
usually have transition metal as coenzyme
73
what does transferase do
transfers function groups form one place to another
74
what do hydrolases do
breaks down covalent bonds in structure of substrate
catalyzes hydrolysis
75
what do lyases do
Cleave C-C, C-O, C-N, or other bonds by elimination, leaving double bonds or rings, or addition of groups to double bonds.
the product of these will always have a double bond present
76
what do isomerases do
involved in central metabolic pathways
transfer of groups withing molecules to yield isomeric forms
77
what do ligases do
formation of covalent bonds by condensation reactions coupled to cleavage of ATP or similar cofactor
they often require ATP to make an exergonic process
78
what do translocases do
move things across membranes or their separation within membranes, they require a conformational change
there are 3 types
79
How do competitive inhibitors bind, how does this inhibitor impact Vmax, Km, Kcat, and efficiency
- compete with the substrate for an active site of an enzyme. The enzyme's catalytic efficiency for the substrate decreases
- by definition the value of alpha' is 1
- when [S]>>[I], the reaction exhibits normal Vmax
- apparent Km decreases by a factor of alpha
-Vmax EQUALs appVmax (unchanged with inhibitor)
- km
80
How do uncompetitive inhibitors bind, how does this inhibitor impact Vmax, Km, Kcat, and efficiency
- binds at site distinct from the substrate active site
- binds ONLY to the ES complex
- Vmax > appVmax (when inhibitor is present Vmax DECREASES)
- km> appkm (when inhibitor is present km DECREASES)
- slopes are equal
-kcat decreases
- catalytic efficiency ??
81
How do mixed inhibitors bind, how does this inhibitor impact Vmax, Km, Kcat, and efficiency
- binds at a site distinct from the substrate active site
- can bind to E or ES complex (why it's mixed)
- Vmax > appVmax (when inhibitor is present Vmax DECREASES)
- km does NOT equal appkm
- presence of the inhibitor INCREASES the slope
- kcat ??
- efficiency??
82
How do noncompetitive inhibitors bind, how does this inhibitor impact Vmax, Km, Kcat, and efficiency
- affects Vmax but NOT Km
- Vmax > appVmax (when the inhibitor is present, Vmax DECREASES)
- Km = appKm (inhibitor has no effect on Km
- The presence of the inhibitor INCREASES the slope
- kcat DECREASES
- catalytic efficiency stays the same
83
Of all the species that enzymes bind, they are thought to bind most tightly to _____.
transition state
84
The highest point in a reaction coordinate diagram represents _____.
transition state
85
An enzyme that forms a covalent bond with its substrate during the course of a reaction is considered to undergo _____.
covalent catalysis
86
Which of the following amino acids would be most likely found in the active site of an enzyme that uses acid-base catalysis?
Asn
Ser
Met
His
Trp
His
87
In acid base catalysis, amino acids that participate in the reaction that converts a ketone to an enol will stabilize the (what) that forms during the (what) state
negative charge
transition
88
What amino acid residue present in the specificity pocket allows trypsin to bind to peptides containing Arg or Lys?
Ser
His
Lys
Val
Asp
Asp
89
The reaction catalyzed by chymotrypsin is an example of what type of reaction?
oxidoreductase
tranferase
hydrolase
lyase
ligase
hydrolase
90
During the first half of the chymotrypsin mechanism where the acyl-enzyme intermediate is formed, what role does His play?
A) general acid only
B) general base only
C) general acid then general base
D) general base then general acid
E) nucleophile
general base then general acid
91
How many transitions states are stabilized during chymotrypsin reaction mechanism
1
0
2
4
2
92
A Lineweaver-Burk plot is a _____.
double reciprocal plot
Michaelis-Menten plot
sigmoidal plot
hyperbolic plot
logarithmic plot
double reciprocal plot
93
When a substrate and enzyme interact, the first chemical species formed is _____.
A) enzyme-substrate complex
B) enzyme-transition state complex
C) enzyme-product complex
D) enzyme plus product
E) none of the above
enzyme-substrate complex
94
How is an enzyme-catalyzed reaction affected by the addition of more enzyme?
A) velocity is not affected
B) velocity will increase only if more substrate is also added
C) velocity will increase
D) velocity will decrease
none of the above
velocity will increase
95
A plot of velocity versus substrate concentration for a simple enzyme-catalyzed reaction produces a _____. This indicates that at some point, the enzyme is _____.
hyperbolic curve; saturated with substrate
96
Which of the following properly expresses the Michaelis-Menten equation?
A) Vo = Vmax [S] / (KM + [S])
B) Vo = Vmax KM / (KM + [S])
C) kcat = Vmax / [E]T
D) Vmax = Vo [S] / (KM + [S])
E) Vmax = Vo KM / (KM + [S])
Vo = Vmax [S] / (KM + [S])
97
Which of the following indicates that an enzyme has evolved to its most efficient form?
A) kcat is a large number
B) KM is a small number
C) KM is a large number
D) kcat/KM is a small number
E) kcat/KM is near the diffusion-controlled limit
kcat/KM is near the diffusion-controlled limit
98
Transition state analogs are generally used as what kind of inhibitors for enzymes?
Competitive inhibitors
Uncompetitive inhibitor
Mixed inhibitor
competitive inhibitors
99
A researcher adds 1M of competitive inhibitor to an existing solution of substrate and enzyme. The researcher notices that the initial velocity of the enzyme decreases. What can the researcher do to increase the velocity of the enzyme back to normal levels (to the level before the inhibitor was added)?
A) Decrease the volume of the solution
B) Increase the volume of the solution
C) Increase the concentration of the substrate
D) Nothing can be done to bring enzyme activity to normal levels
Increase the concentration of the substrate
100
Which of the following is true of uncompetitive inhibitors of enzymes?
A) They bind to both the enzyme-substrate complex and the free enzyme
B) Their effect can be counteracted by adding more substrate
C) They only affect enzymes that act on multiple substrates
D)They decrease the apparent Km and increase the apparent Vmax
E) they lower the concentration of free enzyme available to bind substrate
They only affect enzymes that act on multiple substrates
101
An uncompetitive inhibitor binds to which of the following?
A) An allosteric site on the enzyme only when the substrate has not bound to the active site
B) An allosteric site on the enzyme, only when the substrate is already bound to the active site
C) The active site of the enzyme at the same time as the substrate
D) The active site of the enzyme before the substrate has a chance to bind
E) An allosteric site on the enzyme, regardsless of whenther or not the substrate is already bound to the active site
An allosteric site on the enzyme, only when the substrate is already bound to the active site
102
An unknown inhibitor was dropped into a solution of excess enzyme, and it was found that the enzyme's Km decreased and the Vmax decreased, the change in both was proportional so Km/Vmax remained constant. What kind of inhibition is observed?
uncompetitive
103
Which of the following is true about mixed inhibition?
A) The inhibitor binds independently of the substrate binding
B) The inhibitor binds to the same site as the substrate, dropping the Km
C) The inhibitor competes with the substrate to bind to the active site, and drops the Vmax
D) Vmax stays the same, however Km increases
E) The inhibitor binds to a separate site from the substrate and enhances enzyme activity
The inhibitor binds independently of the substrate binding
104
