MAO isoenzymes and inhibitors
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MAO isoenzymes and inhibitors
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[ID] => 560289
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[post_date] => 2025-01-14 06:48:44
[post_date_gmt] => 2025-01-14 11:48:44
[post_content] => Practice Passage (Question 1-6)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Monoamine oxidase (MAO), comprising isoenzymes MAO-A and MAO-B, is crucial for the oxidative deamination of neurotransmitters such as dopamine, noradrenaline, and serotonin. The isoenzymes, encoded by separate genes on the X-chromosome, are flavoproteins each containing an electron acceptor, flavin–adenine dinucleotide (FAD). The active sites of these enzymes contain key tyrosine residues that help orient the substrate, facilitating its interaction with FAD. Despite sharing structural similarities in their catalytic centers, MAO-A and MAO-B exhibit different intrinsic activities with any given substrate, as shown in Table 1.
Table 1. Kinetic parameters of MAO dependent oxidation of common neurotransmitters
In Parkinson's disease (PD), MAO-B inhibitors like safinamide are used to prevent the breakdown of dopamine, thereby enhancing its availability and alleviating motor symptoms associated with dopamine deficiency. By selectively inhibiting MAO-B, these drugs help manage the progression and symptoms of Parkinson's disease, improving the quality of life for patients. The structures of several MAO inhibitors are shown in Figure 1.
Figure 1. Irreversible (a) and reversible (b) MAO inhibitors
Natural compounds have also been investigated for their MAO inhibitory properties. For instance, curcumin, found in turmeric, and ellagic acid, found in various fruits and berries, have been reported for their potential benefits in managing Parkinson’s symptoms.
Researchers performed in vitro kinetic assays to investigate the mechanism of inhibition of rat brain MAO-B by curcumin and ellagic acid. Both compounds were tested against MAO-B at varying concentrations of benzylamine, a selective substrate. The enzyme kinetics data are presented as double reciprocal Lineweaver–Burk plots in Figures 2 and 3.
Figure 2. Lineweaver–Burk plot of MAO B processing of benzylamine in the presence and absence of varying concentrations of curcumin
Figure 3. Lineweaver–Burk plot of MAO B processing of benzylamine in the presence and absence of varying concentrations of ellagic acid
Adapted from Ramsay RR, Tipton KF. Assessment of Enzyme Inhibition: A Review with Examples from the Development of Monoamine Oxidase and Cholinesterase Inhibitory Drugs, 2017 and Khatri DK, Juvekar AR. Kinetics of Inhibition of Monoamine Oxidase Using Curcumin and Ellagic Acid, 2016
[post_title] => MAO isoenzymes and inhibitors
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[post_modified] => 2025-05-04 07:02:49
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[question] => Which of the following statements is true regarding the affinity of MAO enzymes for the substrates tested, according to the data in Table 1?
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[answer] => 1
[description] => Reason for the Correct Answer:
The Km value is a measure of the affinity of the enzyme for the substrate. Km is equal to the substrate concentration at ½ Vmax.
Enzymes with a high Km have a lower affinity for their substrate (and require more substrate to get to Vmax/2) whereas those with a low Km will have high affinity (and require less substrate to get to Vmax/2).
In the table, you can see that MAO A has the smallest Km if serotonin is its substrate, and MAO B has the smallest Km if phenylmethylamine is its substrate.
Therefore, MAO B has the highest affinity for serotonin and MAO B has the highest affinity for phenylmethylamine.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Of the substrates tested, MAO B exhibits the highest affinity for phenylmethylamine.
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[each_answer] => B. Of the substrates tested, both MAO enzymes exhibit the highest affinity for benzylamine.
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[each_answer] => C. Of the substrates tested, MAO A exhibits the highest affinity for phenylmethylamine.
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[each_answer] => D. Affinity cannot be inferred from the given kinetic parameters.
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => The catalytic efficiency of MAO-A compared to MAO-B is high in the presence of which substrate?
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
An ideal enzyme with high catalytic efficiency would be one that has a very high velocity and a very high affinity for its substrate.
kcat represents the number of substrate molecules converted to product per enzyme molecule per unit of time when the enzyme is fully saturated with substrate. It is often referred to as the enzyme’s turnover number.
Km is a measure of the substrate concentration at which the reaction rate is half of the maximum velocity (Vmax). It provides an indication of the affinity of the enzyme for its substrate; a lower Km value indicates higher affinity.
As such, kcat/Km is an overall measure of catalytic efficiency.
Compared to MAO-B, MAO-A has a much higher kcat/Km and therefore catalytic efficiency when serotonin is the substrate.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Benzylamine
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[each_answer] => B. Serotonin
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[each_answer] => C. Phenylethylamine
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[each_answer] => D. Phenylethylamine and serotonin
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[2] => Array
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[quiz_unique_key] => 2377279144
[question] => The half-life of MAO-B in the brain is 30–40 days. Which of the following statements is NOT ACCURATE with respect to the drug administration in case of PD patients?
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[answer] => 3
[description] => Reason for the Correct Answer:
The half-life of MAO-B in the brain is 30–40 days, meaning that it takes 30–40 days for half of the enzyme molecules to be replaced by new ones.
Reversible inhibitors bind to the enzyme temporarily and can dissociate, meaning their inhibitory effects are not permanent. Frequent dosing is needed to maintain inhibition.
Irreversible inhibitors bind permanently to the enzyme, inactivating it until new enzyme molecules are synthesized. Their effects are long-lasting, and they do not require frequent dosing.
Since reversible inhibitors bind temporarily, frequent dosing is necessary to maintain inhibition. Still, reversible inhibitors are often chosen to avoid the strict dietary restrictions associated with irreversible MAO inhibitors.
Irreversible inhibitors have a long-lasting effect because they permanently inactivate the enzyme. Because irreversible inhibitors bind permanently to MAO, they do NOT need to be administered frequently. The effect lasts until new enzyme is synthesized, which could be 30–40 days for MAO-B.
)
[answers] => Array
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[each_answer] => A. For reversible inhibitory drugs, frequent dosing is required to maintain a high level of inhibition.
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[1] => Array
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[each_answer] => B. Reversible inhibitors of MAO are administered in cases where dietary restrictions are required during the use of irreversible inhibitors of MAO.
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[each_answer] => C. Irreversible drugs have to be administered in regular doses for MAO inhibition.
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[each_answer] => D. The effect of irreversible drugs is long lasting.
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[3] => Array
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[quiz_unique_key] => 1723550837
[question] => The inhibitors Phenelzine, Tranylcypromine and Isatin contain what characteristic structural motifs, respectively?
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[answer] => 2
[description] => Reason for the Correct Answer:
Look for characteristic structural features in Figure 1.
An indole is a bicyclic structure consisting of a benzene ring fused to a pyrrole ring.
A cyclopropylamine is a three-membered cyclopropane ring attached to an amine group.
A phenol is an aromatic organic compound in which a hydroxyl group (-OH) is directly bonded to a benzene ring.
A hydrazine functional group has the formula –NH–NH2.
Accordingly, Phenelzine, Tranylcypromine and Isatin contain a hydrazine, cyclopropylamine and indole, respectively.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Phenol, indole and hydrazine
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[1] => Array
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[each_answer] => B. Hydrazine, cyclopropylamine and indole
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[each_answer] => C. Cyclopropylamine, indole and hydrazine
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[each_answer] => D. Phenol, cyclopropylamine and indole
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[quiz_unique_key] => 2261298308
[question] => The mode of inhibition of curcumin and ellagic acid toward MAO-B is:
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
Lineaweaver Burke plots allow you to view chances in the 1/Km (x-intercept) and 1/Vmax (y-intercept) values caused by an enzyme inhibitor.
The presence of curcumin in Figure 2 did not alter the Km value, but it did decrease the Vmax, as indicated by the decrease in 1/Vmax.
This suggests that curcumin is a non-competitive inhibitor, which does not affect the apparent binding affinity for the substrate but does decrease the maximum velocity of the reaction.
The presence of ellagic acid in Figure 3 did not alter the Vmax value, but it did increase the Vmax, as indicated by the increase in 1/Km.
This suggests that ellagic acid is a competitive inhibitor, which does not affect the Vmax, but which increases the apparent Km of the enzyme and substrate.
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[answers] => Array
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[0] => Array
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[each_answer] => A. uncompetitive for curcumin and mixed for ellegic acid.
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[1] => Array
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[each_answer] => B. noncompetitive for curcumin and competitive for ellagic acid.
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[2] => Array
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[each_answer] => C. noncompetitive for both.
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[each_answer] => D. competitive for both.
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[quiz_unique_key] => 83407773
[question] => What is the type of reaction shown in the figure below, and what are the identities of the missing compounds (1 and 2)?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
The passage states that FAD acts as an electron acceptor in MAO-catalyzed reactions.
This means that FAD should be reduced in the reaction.
Here, you can also see that a carbon compound is being oxidized. Usually gaining bonds to oxygen is associated with oxidation (loss of electrons) in biological reactions.
Therefore, this would be referred to as oxidative deamination of the substrate, in which FAD is reduced to FADH2 and ammonia (NH3) is released.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. Oxidative deamination; 1 = H2O + FADH2; 2 = FAD + NH3
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[1] => Array
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[each_answer] => B. Reductive deamination; 1= NAD+; 2 = NADH
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[2] => Array
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[each_answer] => C. Reductive deamination; 1 = H2O + FAD; 2 = FADH2 + NH3
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[each_answer] => D. Oxidative deamination; 1 = H2O + FAD; 2 = FADH2 + NH3
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[560289|6] => D
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