L14. Enzymes as Drug Targets

Question 1

What happens if a strongly bound substrate gives a product that also binds strongly to the active site?

Answer 1

Enzyme becomes clogged up and is unable to accept any more substrate.

Question 2

If a medicinal chemist knows the position and nature of different binding region with an active site, is it possible to design molecules that will fit that active site, binding strongly, and act as inhibitors?

Answer 2

Yes

Question 3

What is competitive inhibition?

Answer 3

An enzyme regulation method where a substrate can’t bind when a regulatory molecule, having a similar chemical composition to the substrate, binds to that enzyme’s active site. This blocks enzyme activity.

Question 4

Define a reversible inhibitor.

Answer 4

It is a type of competitive inhibitor where the drug is competing with the natural substrate for the active site and is usually Similar in structure to substrate, product or cofactor.

It Binds reversibly to the active site more strongly having additional functional groups to form extra binding interactions to regions that are not occupied by the substrate and Undergoes no reaction.

Question 5

True of False?

Inhibition depends on the strength of inhibitor binding, inhibitor concentration and the time of binding.

Answer 5

True.

Stronger binding, higher concentration, the longer present in the active site = greater inhibition

Question 6

In terms of reversible inhibitors, if the concentration of substrate increases what will happen?

Answer 6

The substrate can compete more effectively with the drug, and so inhibition by the drug will be less effective.

Question 7

Define irreversible inhibitors.

Answer 7

The inhibitor binds irreversibly to the active site by forming covalent bond and the substrate is permanently blocked from the active site.

likely to be similar in structure to the substrate.

Increasing substrate concentration does not reverse inhibition.

Question 8

True or False?

Reversibile inhibitors have toxic side effects.

Answer 8

False. Its irreversible inhibitors.

Question 9

What are the possible toxic side effects of irreversible inhibitors?

Answer 9

Reactive functional groups might react with other proteins, so it is generally better to inhibit with a reversible inhibitor.

Build-up of a particular substrate: Increasing the concentration of substrate will not reverse inhibition as the inhibitors cannot be displaced from the active site.

Question 10

True or False?

Irreversible inhibitors are not competitive inhibitors.

Answer 10

True.

This is because they can’t be removed due to the permenant covalent bonds they form to the enzyme’s active site.

Question 11

Inhibition of lipase will result in _______.

Answer 11

less fat synthesis

Question 12

What is Orlistat?

Answer 12

Orlistat is an anti-obesity drug that inhibits lipase, blocking it from digesting fats in the intestine.

This results in less Fatty acids and glycerol absorption and Leads to reduced biosynthesis of fat in the body.

Question 13

How does lipase aid biosynthesis of fatty acids and glceryol?

Answer 13

It binds between the fatty acid and glycerol backbone of lipid molecules. The digestion products are then absorbed and act as the building blocks for fat biosynthesis in the body.

Question 14

Define allosteric activitation.

Answer 14

Active site becomes available to the substrates when a regulartory molecule binds to a different site on the enzyme.

Positive allosteric modulation (also known as allosteric activation) occurs when the binding of one ligand enhances the attraction between substrate molecules and other binding sites.

Question 15

Define allosteric deactivation.

Answer 15

Negative allosteric modulation (also known as allosteric inhibition) occurs when the binding of one ligand decreases the affinity for substrate at other active sites.

The active site is distorted and is not recognized by the substrate.

For example, when 2,3-BPG binds to an allosteric site on hemoglobin, the affinity for oxygen of all subunits decreases.

Question 16

How do Inhibitors Act at Allosteric Binding Sites?

Answer 16

The Inhibitor binds reversibly to the allosteric binding site via Intermolecular bonds.

The Induced fit alters the shape of the enzyme distorting the Active site and making it unrecognizable to the substrate

• Inhibitor is not similar in structure to the substrate

Question 17

True or False?

Increasing substrate concentration does reverse inhibition of inhibitors bound to allosteric sites.

Answer 17

False.

• Increasing substrate concentration does not reverse inhibition

Question 18

What are Transition State Inhibitors?

Answer 18

Drugs designed to mimic the transition state of an enzyme-catalysed reaction that are likely to bind more strongly than drugs mimicking the substrate or product.

The drug is designed to mimic the stereochemistry and binding properties of the reaction intermediate, but is stable

Question 19

True or False?

Transition states are high energy, transient species and cannot be isolated or synthesised.

Answer 19

True.

This is why Transition State Inhibitors are designed based on reaction intermediates which are closer in character to transition states than substrates or products

Question 20

What is renin?

Answer 20

A Protease that hydrolyzes a specific peptide bond in the protein angiotensinogen to form angiotensin I, a 10 amino acid peptide.

Question 21

What is Angiotensin Coverting Enzyme (ACE)?

Answer 21

a central component of the renin–angiotensin system, which controls blood pressure by regulating the volume of fluids in the body. It converts the hormone angiotensin I to the active vasoconstrictor angiotensin II by removing 2 C-terminal residues from angiotensin I.

Question 22

Explain the renin–angiotensin system. What is it an example of?

Answer 22

Renin, a Protease, hydrolyzes the peptide bond between Leucine and Valine in the protein angiotensinogen, removing Valine and Isoleucine which forms angiotensin I, a 10 amino acid peptide.

Angiotensin Coverting Enzyme (ACE) then removes the 2 C-terminal residues of Histidine and Leucine, forming Angiotensin II. Angiotensin II then Constricts blood vessels causing Retention of fluid in the kidneys, which in turn increases blood pressure.

Question 23

Angiotensin is a 453 amino acid long protein in humans, but only the first ____ amino acids are the most important for its activity.

Answer 23

12

Question 24

How is Renin inhbited?

Answer 24

Transition state inhibition via an aspartic protease in an Acid-base mechanism involving coordination of a water molecule between the two Aspartic Acid amino acids.

1. One Asp activates the water by abstracting a proton, enabling the water to attack the carbonyl carbon of the substrate scissile bond.
2. This forms a tetrahedral intermediate that is then rearranged.
3. Rearrangement of this intermediate leads to protonation of the scissile amide.

Question 25

What is an aspartic protease and how is it formed?

Answer 25

2 conserved Asp residues in the active site for catalytic cleavage of peptide substrates

Question 26

What is Aliskiren?

Answer 26

An orally active non-peptide and the **1st renin inhibitor to be approved by FDA**. It **Binds strongly, using a hydroxyethylene transition-state mimic** that appears similar to the tetrahedral geometry of the reaction intermediate, and is **stable enough to be hydrolysized because it has no leaving group** for the 2nd part of the reaction mechanism.

Question 27

What are suicide inhibitors?

Answer 27

A **special group of irreversible inhibitors** that are **Substrate analogues**. A **very reactive group is generated via normal catalytic action of the enzyme** that **Forms a covalent bond with a nearby functional group within the active site causing irreversible inhibition with very few side effect**s. They are **relatively unreactive until they bind to the active site** of the enzyme.

Question 28

\_\_\_\_\_\_ are responsible for penicilin resistance.

Answer 28

Suicide inhibitors. ## Footnote They catalyze the hydrolysis of the penicillin β-lactam ring

Question 29

Explain the Reaction catalyzed by bacterial β-lactamase enzymes.

Answer 29

Serine reacts with Penicilin to open penicilin's ring and form the nucleophile that is then reacted with β-lactamase and H+ to form an intermediate where Serine is covalently linked via an ester group to the ring-opened penicillin. β-lactamase and water react with the intermediate to hydrolyze the ester group and release the inactivated penicillin and free up the active site, such that the catalytic process can be repeated.

Question 30

True or False? Transition-state analogues react irreversibly with the enzyme.

Answer 30

False

Question 31

\_\_\_\_\_\_\_\_\_ require the presence of a stable functional group to mimic the functionality present in the transition state.

Answer 31

Transition state inhibitors (analogues)

Question 32

True or False? Suicide substrates self-destruct as a result of an enzyme-catalyzed reaction.

Answer 32

False. ENZYMES SELF DESTRUCTION NOT INHIBITORS: THESE ARE CALLED “SUICIDE” BECAUSE THEY LEAVE ON THEIR OWN

Question 33

Which amino acids would be prone to a reaction with an irreversible inhibitor?

Answer 33

Serine

Question 34

During the absorption of dietary fat, the molecule is broken down before being reassembled following absorption. What is dietary fat broken down into before absorption? A. Three fatty acids and one monoglyceride B. Two fatty acids and one monoglyceride C. Triacylglycerol D. Triglycerides

Answer 34

B. Two fatty acids and one monoglyceride **Note** that triglycerides are dietary fat, which are also known as triacylglycerol, and neither are therefore breakdown products.

Question 35

Dietary fat is broken into two fatty acids and one monoglyceride prior to absorption. Due to the activity of lipase, triglycerides are broken down into two fatty acids and a monoglyceride before absorption and \_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer 35

Ultimately reassembled into a triglyceride in the enterocyte

Question 36

A biochemist assesses the changes in reaction rate of an enzyme-catalyzed reaction in a solution containing saturating quantities of a known inhibitor. When substrate is added, the reaction rate increases, but fails to reach the published Vmax for this enzyme concentration, even at concentrations of substrate well above that of the inhibitor. Interestingly, the researcher notes that the concentration of substrate required to reach half of the observed maximal rate matches the published value. What is **true** of this enzyme and its inhibitor? A. The inhibitor binds to active site of the enzyme. B. Inhibitor concentration has no impact on reaction rate. C. Addition of inhibitor does not impact enzyme affinity for the substrate. D. The Vmax is independent of enzyme concentration.

Answer 36

C. Addition of inhibitor does not impact enzyme affinity for the substrate. The stem of the question describes a situation where an enzyme is in the presence of a saturating quantity of inhibitor and substrate is added. Given we have a decreased Vmax with an unchanged Km, this is likely a noncompetitive inhibitor.

Question 37

Methoxy arachidonyl fluorophosphates (MAFPs) covalently link to the active sites of serine proteases. What best characterizes the activity of a serine protease following MAFP binding? A. Competition between the substrate and the MAFP which can be overcome with high substrate concentrations B. Increased protease activity C. Negligible protease activity D. Activation of the protease activity of the serine protease to cleave the MAFP

Answer 37

C. Negligible protease activity Here, a serine protease is treated with an MAFP that covalently binds to the active site. This would result in an irreversible loss of protease activity and no amount of substrate could displace the covalently linked MAFP (competition is not possible when the inhibitor binds covalently).

Question 38

Which of the following is an example of feedback inhibition? A. Decreased activity of hexokinase due to a decreased concentration of its substrate glucose B. Decreased activity of hexokinase due to elevated levels of a competitive inhibitor, 2-deoxyglucose C. Decreased activity of PFK due to elevated levels of phosphoenolpyruvate, a glycolytic intermediate D. Increased activity of PFK due to high levels of AMP

Answer 38

C. Decreased activity of PFK due to elevated levels of phosphoenolpyruvate, a glycolytic intermediate. Feedback inhibition involves the decrease in activity of an enzyme **by binding to a downstream product**. PFK, the rate-limiting enzyme in glycolysis, is inhibited by **phosphoenolpyruvate (PEP) which is an intermediate in the glycolytic pathway**. Hexokinase is the first enzyme in the pathway which phosphorylates glucose. Decreasing concentrations of glucose may limit the activity of hexokinase, **but as it is a substrate, it would not be characterized as feedback inhibition**.

Question 39

Which of the following is **NOT** a known mechanism of direct enzymatic regulation? A. Interaction of the enzyme with downstream products B. Peptide hydrolysis C. Removal of a phosphate with the use of a phosphatase D. Varied activity of transcription factors affecting enzymatic expression

Answer 39

**D. Varied activity of transcription factors affecting enzymatic expression.** While varying activity of transcription factors will affect enzymatic expression, it is an indirect method of regulation. all other answers are DIRECT enzymatic regulation

Question 40

Which of the following would **most affect the Km** of an enzyme? A. A mutation in a highly variable region of the protein coding sequence B. Addition of a noncompetitive inhibitor C. Increasing substrate concentration D. A mutation in the active site

Answer 40

**D. A mutation in the active site** The Michaelis constant **(Km) is the concentration of substrate necessary for a reaction to achieve ½ Vmax and is** **_inversely proportional to affinity of the enzyme for its substrate_**. Therefore a mutation in the active site of the enzyme is likely to affect the Km for a given reaction.

Question 41

Which of the following would be **true** of the Lineweaver-Burk plot for a non-competitive inhibitor? A. The y-intercept of the graph would be bigger and the x-intercept would be smaller. B. The y-intercept of the graph would be bigger, and the x-intercept would not change. C. The y-intercept of the graph would not change and the x-intercept would be bigger. D. The y-intercept of the graph would be smaller and the x-intercept would not change.

Answer 41

**B. The y-intercept of the graph would be bigger, and the x-intercept would not change.** The y-intercept represents Vmax; in non-competitive inhibition, Vmax is reduced, so on an inverse plot this would appear as a bigger y-intercept. However, Km in non-competitive inhibition stays the same; Km is represented by the x-intercept, so the x-intercept would not change.

Question 42

Considering the equilibrium between the free enzyme and enzyme-substrate complex, what would be the expected effect on this equilibrium due to the addition of an uncompetitive inhibitor? A. A shift to the right due to decreased [ES] B. A shift to the left due to decreased [ES] C. A shift to the right due to increased [ES] D. A shift to the left due to increased [ES]

Answer 42

**A. A shift to the right due to decreased [ES]** uncompetitive inhibitor binds to the enzyme-substrate complex, effectively decreasing its overall concentration. **According to Le Chatelier’s principle, a decrease in the concentration of a compound will lead to a shift to form more of that compound**

Question 43

How does uncompetitive inhibition affect the apparent Km and the Vmax of a given enzyme?

Answer 43

Decreased apparent Km and Vmax; both are affected equally