Question 1
Question
Enzymes alter the position of equilibrium of a reaction
Question 2
Question
Reactions are energetically favourable when...
Question 3
Question
If a compound exists in two stereoisomer forms, an enzyme can act upon both of them.
Question 4
Question
Substrate specificity is the notion that enzymes [blank_start]catalyse[blank_end] one type of reaction dependent on substrate/active site [blank_start]complementarity[blank_end]. This means that multiple enzymes may be required to produce a particular product from a single substrate. Thus enzymes [blank_start]compartmentalise[blank_end] in cells.
Answer
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catalyse
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complementarity
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compartmentalise
Question 5
Question
What is the definition of an oxidoreductase?
Answer
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Catalyses transfer of hydrogen atoms and electrons in redox reactions
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Catalyses transfer of functional groups from donors to acceptors
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Catalyses cleavage of C-C/C-O/C-N bonds by adding groups to double bonds or removing groups to form double bonds
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Catalyse cleavage of bonds by addition of water in hydrolysis
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Catalyse transfer of functional groups within same molecule
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Catalyse formation of new covalent bonds using ATP
Question 6
Question
What is the definition of a transferase enzyme?
Answer
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Transfers functional groups from one position on a molecule to another
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Transfers functional groups from donors to acceptors
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Transfers electrons from donors to acceptors
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Transfers an anchored protein from one position on the cell membrane to another
Question 7
Question
What is the definition of a hydrolase enzyme?
Answer
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Catalyses breakdown of hydrogen peroxide to oxygen and water
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Catalyses cleavage of bonds by addition of water
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Catalyses formation of water from oxygen and hydrogen
Question 8
Question
Which type of bonds do ligases catalyse the formation of using energy from ATP?
Question 9
Question
The lock and key model of enzyme action states that the active site has a [blank_start]specific[blank_end] shape [blank_start]independent[blank_end] to that enzyme. The substrate is exactly [blank_start]complementary[blank_end] to the [blank_start]shape[blank_end] of that active site.
Answer
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specific
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independent
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complementary
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shape
Question 10
Question
The modified lock and key model states that there is [blank_start]general[blank_end] complementarity between the enzyme and the substrate but not exact. As the substrate [blank_start]binds[blank_end] to the active site, the [blank_start]enzyme[blank_end] and [blank_start]substrate[blank_end] distort to become complementary. This increases the [blank_start]energy level[blank_end] of the substrate and reduces the [blank_start]transition energy[blank_end] required for the reaction.
Answer
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general
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binds
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enzyme
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substrate
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energy level
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transition energy
Question 11
Question
In the induced fit model of enzyme action, what distorts to provide enzyme-substrate complementarity?
Question 12
Question
What is a cofactor?
Answer
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A species that contributes to the normal functioning of enzymes
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A species that binds to actin filaments and integrin proteins in the cytoskeleton
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A species that binds to haemoglobin to increase its affinity for oxygen
Question 13
Question
What are isoenzymes?
Question 14
Question
In the catalysis of peptide bond hydrolysis by chymotrypsin, the polypeptide bind [blank_start]non-covalently[blank_end] with side chains in the hydrophobic pocket. A [blank_start]proton[blank_end] is tranferred from serine to histidine in the active site. This forms a [blank_start]tetrahedral[blank_end] transition state between the enzyme and the substrate. The [blank_start]C=O[blank_end] group in an amide bond of the substrate forms a single bond to the [blank_start]oxygen[blank_end] in serine. The [blank_start]proton[blank_end] is then transferred from histidine to the [blank_start]C-terminal[blank_end] fragment. This fragment is then released by cleavage of the [blank_start]C-N[blank_end] bond. A [blank_start]water[blank_end] molecule then binds to the histidine in place of the C-terminal fragment. The water molecule transfers a [blank_start]proton[blank_end] to histidine and the remaining [blank_start]hydroxide[blank_end] binds to the carbon of the N terminal fragment. [blank_start]A new tetrahedral[blank_end] transition state is formed. Cleavage of the [blank_start]acyl[blank_end] bond releases the N terminal fragment. Histidine's [blank_start]proton[blank_end] is transferred back to serine and the enzyme regains it's original structure.
Answer
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non-covalently
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proton
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tetrahedral
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C=O
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oxygen
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proton
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C-terminal
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C-N
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water
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proton
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hydroxide
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A new tetrahedral
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acyl
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proton
Question 15
Question
At low substrate concentration, what is the order of reaction for an enzyme-catalysed reaction with respect to substrate concentration?
Question 16
Question
At high substrate concentration, what is the order of reaction of an enzyme-catalysed reaction with respect to substrate concentration?
Question 17
Question
K1 is the rate constant for the [blank_start]forward reaction[blank_end]. This is the [blank_start]formation of ES[blank_end].
K-1 is the rate constant for the [blank_start]reverse reaction[blank_end]. This is the [blank_start]breakdown of ES to E+S without catalysis[blank_end].
K2 is the rate constant for the [blank_start]breakdown of ES into E + P[blank_end].
Answer
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forward reaction
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reverse reaction
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second reaction
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formation of ES
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breakdown of ES to E+S without catalysis
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breakdown of ES into E + P
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reverse reaction
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forward reaction
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second reaction
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breakdown of ES to E+S without catalysis
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formation of ES
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breakdown of ES into E + P
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breakdown of ES into E + P
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formation of ES
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breakdown of ES to E+S without catalysis
Question 18
Question
In the Michaelis-Menten reaction model of enzyme kinetics, we use K-2 as the rate constant for the breakdown of the enzyme-product complex to the enzyme-substrate complex.
Question 19
Question
What are the assumptions of the Michaelis-Menten model of enzyme kinetics? Check all that apply
Answer
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[S] >> [E] so a relatively small amount of substrate is bound to the enzyme at one time
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[ES] does not change with time because rate of formation is equal to rate of breakdown
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All velocities are initial velocities taken from the point of mixture
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[E] is always infinite
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The reaction occurs at pH 7
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The reaction occurs at room temperature
Question 20
Question
[blank_start]V0[blank_end] is the initial reaction velocity - the rate of reaction at the point of enzyme-substrate mixture.
[blank_start]Vmax[blank_end] is the maximal velocity when all enzyme active sites are saturated.
[blank_start]Km[blank_end] is the Michaelis constant.
Answer
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V0
-
Vmax
-
Km
-
Vmax
-
V0
-
Km
-
Km
-
V0
-
Vmax
Question 21
Question
The Michaelis constant is a measure of...
Answer
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The affinity of the enzyme for the substrate
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The catalytic efficiency of the reaction
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The turnover number of the enzyme
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The number of enzymes saturated with substrate at any one time
Question 22
Question
The lower the value of Km, the higher the affinity of the enzyme for the substrate.
Question 23
Question
What equation is this for?
Answer
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Initial velocity of an enzyme-catalysed reaction
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Turnover number of an enzyme in a particular reaction
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The Michaelis constant
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The catalytic efficiency of an enzyme-catalysed reaction
Question 24
Question
What is the equation for Km, the Michaelis constant?
Answer
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(k-1 + k2) / k1
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(k1 + k2) / k-1
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(k-1 + k1) / k2
Question 25
Question
Km = [S] when...
Answer
-
V0 = Vmax/2
-
V0 = Vmax/4
-
V0 = Vmax
Question 26
Question
What denotes the equilibrium constant for the formation of ES in terms of Michaelis-Menten kinetics?
Answer
-
k-1 / k1
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k2 / k1
-
k1 / k2
-
k1 / k-1
Question 27
Question
If k2 is very small, the rate of reaction is...
Question 28
Question
What is the name given to the number of substrate molecules converted to product in a given unit of time on one enzyme molecule when enzyme is saturated with substrate?
Answer
-
Turnover number
-
Product number
-
Catalysis number
Question 29
Question
The higher the Kcat number, the faster the rate of reaction
Question 30
Question
The specificity constant should be high to give a high efficiency of an enzyme-catalysed reaction. Therefore...
Answer
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Kcat should be high and Km should be low
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Km should be high and Kcat should be low
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Kcat and Km should be equal
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Kcat should be negative
Question 31
Question
Lactate dehydrogenase consists of [blank_start]4[blank_end] monomers of either heart or [blank_start]muscle[blank_end] type. This allows the existence of [blank_start]5[blank_end] different isozymes. Lactate dehydrogenase catalyses the conversion of lactate to [blank_start]pyruvate[blank_end]. In the leg muscles there is a [blank_start]fast[blank_end] rate of lactate formation. Therefore, we need an enzyme with high [blank_start]sequestering[blank_end] ability. This means that lactate dehydrogenase in the leg muscles requires a [blank_start]low[blank_end] Km number. In the heart, lactate is toxic if left free so must be removed as quickly as possible. This means that lactate dehydrogenase in the heart requires a high [blank_start]Kcat[blank_end] number. During a [blank_start]myocardial infarction[blank_end], rupture of the heart endothelial cells causes lactate dehydrogenase enzymes to increase in levels in serum. These levels can therefore be used as a diagnostic tool.
Answer
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4
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muscle
-
5
-
pyruvate
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fast
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sequestering
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low
-
Kcat
-
myocardial infarction
Question 32
Question
What is the correct rearrangement of the Michaelis-Menten equation to give the Lineweaver-Burke equation?
Answer
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1/V0 = Km/Vmax x [S] + 1/Vmax
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1/V0 = Vmax/Km x [S] + 1/Vmax
-
V0 = Km/Vmax x [S] + 1/Vmax
-
V0 = [S]/Vmax x Km + [S]
Question 33
Question
1/Km on a Lineweaver-Burke plot is found by the x intercept. This value can be positive or negative.
Question 34
Question
A competitive inhibitor will alter the apparent Km value of an enzyme but not Vmax.
Question 35
Question
What will happen to the y intercept on the Lineweaver-Burke plot of a reaction taking place in the presence of a competitive inhibitor compared to the regular reaction?
Answer
-
Decreases
-
Increases
-
Doesn't change
Question 36
Question
What will happen to the x intercept on a Lineweaver-Burke plot of an enzyme-catalysed reaction in the presence of competitive inhibitor?
Answer
-
Decreases
-
Increases
-
Stays the same
Question 37
Question
Non-competitive inhibitors do not alter the Km of an enzyme.
Question 38
Question
What happens to the gradient of a Lineweaver-Burk plot in the presence of a non-competitive inhibitor?
Answer
-
Increases
-
Decreases
-
Stays the same
Question 39
Question
What is an allosteric binding site?
Answer
-
A site other than the active site
-
The active sites of enzymes
-
The site where a signal binds to a receptor protein
-
The site where an enzyme binds to DNA
Question 40
Question
What is the name given to a molecule that binds to the site of an allosteric enzyme, causing a change in activity?
Answer
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Allosteric effector
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Inhibitor
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Stabiliser molecule
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G-protein
Question 41
Question
Which residues can phosphate be added or removed from? Check all that apply
Answer
-
Serine
-
Thymine
-
Threonine
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Histidine
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Cysteine
-
Alanine
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Arginine
-
Glutamic acid
Question 42
Question
Phosphorylation increases the activity of enzymes