L6-8: Intro to Enzymes I, II & III Flashcards

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1
Q

What are activating subunits?

A
  • these are components of a holoenzyme that are required for the enzyme’s (of which they are associated) catalytic activity
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2
Q

Draw the Lineweaver-Burk plot and indicate what each intercept, slope and axes mean. Indicate what navigating around the graph means.

A
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2
Q

What are two common competitive inhibitor drugs used in everyday life?

A
  • Acetaminophen and advil. These are prostaglandin synthase inhibitors.
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3
Q

Do enzymes change the free energy of the substrates and / or products of a reaction?

A
  • No, delta G is unchanged.
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3
Q

An enzyme has a histidine residue that is essential for catalysis. This enzyme is inactive a pH < 4 and pH > 9. Why?

A
  • Histidine residues have a pKa of approximately 6.0, which at physiological pH means that they can exist in protonated or deprotonated form. When at pH of 4, residue has strong tendency to exist in protonated form (acid) and cannot act as a good proton donor. When at pH of 9, residue has strong tendency to exist in deprotonate form (base) and cannot act as a good proton acceptor. Therefore, catalysis relying on acid/base interactions between enzyme and substrate no longer occur and function is lost. All residues have optimum pH range in which they work to enhance reaction rates, extremes and out of this optimum have decreased or negligible reaction rates.
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3
Q

How can one experimentally distinguish between noncompetitive and irreversible inhibition?

A
  • removal of irreversible inhibitor does not restore enzyme activity; however, removal of non-competitive inhibitor will restore enzyme activity.
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3
Q

How does aspirin function?

A
  • It is an irreversible inhibitor of prostaglandin synthase and binds to the active site of the enzyme forming a covalent complex.
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4
Q

Define cofactors.

A
  • Cofactors are defined as small molecules that are required for activity of enzymes. They include metals (metaloenzymes) and small organic molecules (coenzymes).
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5
Q

The Km for a substrate often corresponds closely to metabolic concentrations of the substrate. Why is this advantageous?

A
  • When changes to substrate concentration occur, enzyme is sensitive enough to decrease reaction rate or increase reaction rate.
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6
Q

Late onset hyperammonemia results in elevated ammonia in the blood and is caused by a mutation to an arginine residue to be replaced by a glutamine residue. As a result, Km is 60 fold higher than the normal enzyme. Explain how hyperammonemia is occurring.

A
  • Since the mutated protein has a Km that is 60 fold higher than the normal protein, this means that the protein has low affinity for the substrate. At physiological substrate concentrations, reaction rates are now lower due to decreased enzyme-substrate interactions (ez activity) and necessary down stream reactions that serve to “rid” the body of excess ammonia and hindered.
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6
Q

Explain what an irreversible inhibitor is?

A
  • An irreversible inhibitor is one that forms covalent complexes with active site residues in enzymes, chemically modifying and therefore inactivating an enzyme. No change occurs with increased substrate concentration, nor removal of inhibitor.
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8
Q

What are active sites? Where are active sites often located? Why?

A
  • Binding sites refer to small areas in an enzyme (relative to total enzyme structure) that bind substrate. They are often in a pocket, crevice or cleft in the enzyme where water is excluded (unless participating in reaction). They need to be discrete and hidden to prevent cross reactions / side reactions from occurring – specificity is key.
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8
Q

What are suicide substrates?

A
  • Suicide substrates, aka Trojan Horse substrates, are special classes of irreversible inhibitors that only become inhibitors through the catalytic action of the target enzyme. They bind covalently at active site residues.
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9
Q

Describe how enzymes can react with chiral substrates in a stereospecific manner.

A
  • In order for an enzyme to bind a chiral substrate, there needs to be 3 points of interaction between the substrate and the enzyme. For this purpose, if only two atoms branched from a chiral center interact with the enzyme, but not the 3rd because the substrate is of a different stereochemistry, a reaction will not occur. Needs 3 point interaction. This allows for stereospecific binding and reactions.
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9
Q

How do you convert the rectangular hyperbolic V vs. [S] curve into a straight line using the Lineweaver-Burk (double-reciprocal) method. What are the X- and Y-intercepts and the slopes of this plot? How could the Vmax and Km be determined using this plot?

A
  • 1/v = (Km/[S]).(1/Vmax) + 1/Vmax (y = mx + b form) - The x intercept = -1/Km – the negative reciprocal gives you Km - The y-intercept = 1/Vmax – the reciprocal gives you Vmax - Slope = Km/Vmax
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9
Q

How do many drugs work against enzymes?

A
  • Many drugs reduce the rate of ez catalyzed reactions. Most drugs are reversible, competitive inhibitors, meaning they compete with substrate for binding at the active site.
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10
Q

Why are non-competitive inhibitors potentially better drugs than competitive inhibitors? What is the difficulty in manufacturing them?

A
  • Non-competitive inhibitors would perhaps make better drugs since they can inhibit enzyme activity irrespective of the substrate concentration. Would not have to worry about dosing based on substrate concentration. They are difficult to manufacture from the design perspective. Since non-competitive inhibitors bind in a location other than the active site, yet upon binding cause active site to be unable to bind substrate, how can you determine where this other site is?
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12
Q

Define coenzymes.

A
  • Coenzymes are cofactors that are small organic molecules. Coenzymes are typically derived from vitamins. Eg. B1, niacin, B6, B2 etc.
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13
Q

Describe how the determination of the 3-D structures of the isozymes of prostaglandin synthase aided in the design of drugs that selectively inhibit COX-2.

A
  • NSAIDs act by inhibiting prostaglandin synthase, which are responsible for pain and inflammation. This enzyme has two enzymatic activities, one of which is cyclooxygenase activity and the other, hydroperoxidase. NSIADs inhibit the cyclooxygenase. Turns out there are two isozymes for this enzyme, COX-1 and 2. - COX-1 protects gastrointestinal mucosa, while COX-2 is associated with inflammation of arthritis. - As a result of this discovery, drugs have been designed specifically as inhibitors to COX-2. Eg. Celebrex, bextra and vioxx.
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13
Q

What is product inhibition? Give an example of an enzyme that is regulated by product inhibition.

A
  • Refers to the product of an enzyme catalyzed reaction inhibiting the enzyme that made it when concentrations of itself are high. - Example: G6P inhibits hexokinase
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14
Q

What are the advantages of a multienzyme/multifunctional protein complex?

A
  • Elimination of substrate diffusion/dissociation between sequential enzymatic reactions. - Coordinated control of sequence enzymatic steps. - Coordinated, stoichiometric gene expression of enzymatic activities.
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15
Q

What is the relationship between the magnitude of Km and enzyme/substrate affinity?

A
  • inverse relationship - low Km = high S-E affinity - high Km = low S-E affinity - means some enzyme-substrates interact better than others
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16
Q

Describe and explain graphically with 1/V vs. 1/[S] plots and pictorially with drawings: • competitive inhibition

A
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18
Q

Define metalloenzymes.

A
  • Metaloenzymes are enzymes with necessary metallic ions that allow the enzyme to be active and function. Eg. Fe, Cu, Mo, Zn, Mg etc.
19
Q

Explain how interactions of an enzyme with its substrates accelerate biochemical reactions in terms of: • proximity and orientation • acid-base catalysis • covalent catalysis • metal ion catalysis

A
  • In order for a reaction to proceed, the number collision where the geometry is correct for transition state formation is low. Enzymes facilitate formation of the transition state by binding substrates in their correct molecular orientation for transition state complexes to occur. Therefore energy of activation for a reaction to proceed is lowered and increased reaction rates occur. - Enzymes can act as biological catalysts by allow their R-groups in the active site to serve as proton donors/acceptors depending. Depending on the pKa of the R-group in the context of the pH of the envionment, it will either be more willing to be a donor/acceptor. pH pKa: base predominates – proton off – acts as proton acceptor. - R-groups within active sites of enzymes are able to become strong nucleophiles and as a result transiently form covalent bonds between residue and substrate then further proceed to free enzyme with product as reaction proceeds. - Metal ions cofactors can participate in catalysis by: a.) positive charge can stabilize negatively charged intermediates; b.) positive charge can generate a nucleophile by increasing acidity of nearby molecule (water) and c.) metal ion can bind to substrate increasing number of enzyme-substrate interactions
21
Q

Describe the interaction of the substrate and active site as proposed by the induced-fit model.

A
  • The induced-fit model states that when binding of substrate occurs, it induces a conformation change in the active site.
22
Q

List three types of regulatory subunit that can interact with enzyme catalytic subunits. Contrast the mechanisms by which these subunits function.

A
  • a.) inhibitory subunits: these are components of a holoenzyme that inhibit catalytic activity of the enzyme they are associated with. - b.) activating subunits: these are components of a holoenzyme that are required for the enzyme’s (of which they are associated) catalytic activity - c.) targeting subunits: these are molecules that bind to and direct catalytic subunits on enzymes to their substrates in the cell. They are not components of a holoenzyme.
23
Q

Explain what a competitive inhibitor is?

A
  • A competitive inhibitor is a reversible inhibitor. These compete with substrate to bind at the active site on the enzyme. Increasing substrate concentration will prevent inhibitor binding. Infinite concentration of substrate will abolish inhibition. Inhibitor once removed will not affect enzyme functionality. - These are the most common type of drug
24
Q

What is an irreversible inhibitor? Give a common example.

A
  • Irreversible inhibitors chemically modify and inactivate enzymes by forming covalent complexes at the active site residues. Aspirin is an example and it is a prostaglandin synthase inhibitor.
25
Q

What are allosteric modulators? Do enzymes that are regulated by allosteric modulators follow Michaelis-Menten kinetics?

A
  • Allosteric modulators are small molecules that regulate enzyme activity by binding to sites distinct from the active sites. Enzymes that are allosteric, do not follow Michaelis-Menten kinetics.
27
Q

What does U mean?

A
  • U is the international unit, which expresses v, or the rate of an enzyme catalyzed reaction. - Is the is amount of enzyme required to convert 1 umol of substrate into product in 1 minute at saturating substrate concentration.
28
Q

What simple step can the body take to make enzyme catalyzed reactions go faster?

A
  • make more enzyme
29
Q

What are zymogens? Describe how chymotrypsinogen is activated in the digestive tract.

A
  • Zymogens are inactive enzyme precursors that become activated through proteolysis in order to function. This is an irreversible process. - Chymotrypsinogen is an inactive precursor that is cleaved by trypsin to pi-chymotrypsin (which is an active form) and then self-cleaves to alpha-chymotrypsin (also an active form).
31
Q

Describe and identify the reactions catalyzed by the following classes of enzymes: oxidoreductases, transferases, hydrolases, lyases, isomerases and ligases.

A
  • a.) oxidoreductases: class I: catalyze redox reactions, involve electron acceptors / donors incl. oxidases, oxygenases, reductases, dehydrogenases - b.) transferases: class II: transfer groups (carboxyl, amino, glucosyl, phosphoryl, methyl etc.) from one substrate to another without the input of energy incl. kinases, aminotransferases, carboxylases, methyltransferases - c.) hydrolases: class III: cleave bonds by addition of water incl. glucosidases, ATPases, phosphatases, peptidases, lipases - d.) lyases: class IV: break bonds without addition of water or oxidative cleavage - e.) isomerases: class V: change steoreochemistry of optical or geometric isomers incl. epimerases, mutases, racemases - f.) ligases: class VI: catalyze formation of bonds via input of energy (usually via hydrolysis of XTP), not via transfer to bond incl. ligases, synthases, synthetases, carboxylases
32
Q

How do you increase the rate of a reaction once enzyme saturation has occurred?

A
  • Add more enzyme to the mix. It will increase proportionally to the amount of enzyme added (provided substrate is available). Double the concentration of enzyme, reaction rate doubles.
33
Q

Distinguish between multienzyme complexes and multifunctional proteins.

A
  • Multienzyme complexes are formed by the association of protein subunits, each of which is a distinct enzyme. This can be the result of 10s, 20s of distinct enzymes. Include proteasomes.
35
Q

Are carboxylases ligases or transferases?

A
  • Depends if a group is transferred or formed. Carboxylase is a ligase if it catalyses the formation of bonds with required input of energy via hydrolysis of XTP. It is a transferase if it transfers a group from one substrate to another without input of energy.
36
Q

The Km of an enzyme is independent of enzyme concentration while the velocity of the reaction (including Vmax) is directly proportional to enzyme concentration. Why?

A
  • Km reflects the substrate concentration at which v = Vmax / 2. Therefore, when changing the enzyme concentration, the rate will change in proportion to that. What is also true, is that increase substrate concentration will increase reaction rate to Vmax. Km is calculated by looking at rate constants, which are independent on enzyme concentration and mere just a reflection of the properties of the enzyme. No matter how much enzyme you have, substrate interaction with the enzyme are limited by its intrinsic properties.
37
Q

How do enzymes affect: • the energy of activation for the forward and reverse reaction? • the equilibrium constant of a reaction? • the equilibrium concentrations of reactants and products? • the forward and reverse rate constants for the reaction?

A
  • Ea is reduced by the same amount for the forward and reverse reactions - Equilibrium constant (Keq) is unchanged as energy of reactants and products is unchanged - Equilibrium concentrations of reactants and products are unchanged - Forward and reverse rate constants are accelerated by same amount
39
Q

Define apoenzyme.

A
  • Are enzymes from which a prosthetic group has been removed. Eg. Hb without heme
40
Q

Describe the clinical importance of LDH isozyme profiles.

A
  • There are 5 isozymes of LDH and they are distributed throughout the body in different abundances. - The normal LDH isozyme profile looks at relative abundances of each LDH isozyme and indicates that LDH2 is highest. - After an MI; however, this profile changes and LDH1 (found mostly in the heart and RBCs) becomes elevated. This used to be an important diagnostic tool to determine heart attacks. If LDH5 is evelated, it is diagnostic for acute hepatitis.
41
Q

What are two major classes and subclasses of drugs?

A

1.) Reversible inhibitors: removal of inhibitor fully restores enzyme activity a.) competitive: compete with and bind at substrate’s active site b.) non-competitive: binds at site other than active site forming inactive enzyme 2.) Irreversible inhibitors: removal of inhibitor does not restore enzyme activity

42
Q

Many enzymes are regulated via reversible phosphorylation. What families of enzyme are responsible for (a) the addition and (b) the removal of phosphate? What three amino acid residues can be phosphorylated?

A
  • Kinases are responsible for the addition of phosphate groups and phosphatases are responsible for the removal of phosphate groups. Serine, threonine and tyrosine are the AA residues that have the ability to be phosphorylated.
43
Q

Why know Km? What does it tell us?

A
  • Km approximates the dissociation constant (Kd) of th ez-substrate complex. - Low Km = high substrate affinity = tight binding - High Km = low substrate affinity = weak binding
45
Q

A gradual increase in temperature will increase the rate of an enzyme catalyzed reaction. However as the temperature continues to increase, the enzyme suddenly loses catalytic activity. Why?

A
  • As temperature increases, speed at which molecules move into the active site increase and therefore catalysis increases. As the temperature continues to increase; however, the enzyme becomes denatured through disruption of non-covalent interactions (such as H bonding, salt bridges, VDW forces, hydrophobic interactions). Structure of enzyme is lost, therefore function is lost.
45
Q

Describe and explain graphically with 1/V vs. 1/[S] plots and pictorially with drawings: • noncompetitive inhibition

A
46
Q

What are inhibitory subunits?

A
  • these are components of a holoenzyme that inhibit catalytic activity of the enzyme they are associated with.
47
Q

What are targeting subunits?

A
  • these are molecules that bind to and direct catalytic subunits on enzymes to their substrates in the cell. They are not components of a holoenzyme.
49
Q

What are isoenzymes? What is the physiological significance of the difference in Km for glucose between hexokinase and glucokinase? Which enzyme has the higher affinity for glucose? Physiological implication?

A
  • Isoenzymes are enzymes with different amino acid sequences, but catalyze the same biochemical reaction. They are distributed differently between various tissue and have different physical properties (Km, vmax, etc.) - Glucokinase and hexokinase are enzymes that convert glucose to G6P. Glucokinase is found in liver and hexokinase is found in all other cells. Km of glucokinase for glucose is 5mM, while Km of hexokinase is 0.1 mM. Hexokinase has low Km, high affinity and low Vmax. Glucokinase has high Km, low affinity and high Vmax. This means that when blood glucose levels continue to rise, hexokinase is running at Vmax becomes saturated and cannot hold onto glucose anymore, glucokinase will still be able to respond to this increased/high concentration.
50
Q

Write the Michaelis-Menten equation. Define what each of the variables are.

A
  • v = (Vmax [S])/(Km + [S]) - v = rate of ez catalyzed reaction - Vmax = maximum reaction rate (reached when every ez molecule is bound to substrate). Depends only on ez concention. - [S] = substrate concentration - Km = Michaelis-Menten constant (k2+k3)/k1 – it is the substrate concentration at which v=Vmax / 2 . Km is independent of enzyme concentration, reflects physical properties of enzyme.
51
Q

What are the 6 classes of enzymes?

A
  • Oxidoreductases, transferases, hydrolase, lyases, isomerases and ligases
52
Q

Draw a substrate saturation curve and indicate Vmax, Km, Vmax/2.

A
53
Q

Define holoenzyme.

A
  • Holoenzyme refers to the apoenzyme plus its prosthetic group. Eg. Hb with heme = holoenzyme. - Also refers enzymes with distinct protein subunits.
54
Q

Explain what a non-competitive inhibitor is?

A
  • A non-competitive inhibitor is a reversible inhibitor. These bind at sites on the enzyme other than the active site, forming inactive enzymes (E-I form or E-S-I). Increasing the concentration of the substrate does not influence inhibitor binding. It is reversible if the inhibitor can be removed. Enzyme will work after inhibitor removal.
55
Q

Define prosthetic groups.

A
  • Prosthetic groups are coenzymes that are attached to their enzymes. Eg. Heme is prosthetic group in Hb/Mb.
56
Q

Calculate v: If the Km is 5 mM, what is the reaction rate at 20 mM? Express the rate as a percentage or fraction of Vmax.

A

v = Vmax [20 mM] / (5 + 20 mM) = 20Vmax/25 = 4/5 Vmax

57
Q

Calculate v: If the substrate concentration is 10 mM, the Km is 40 mM and the Vmax is 100 U, what is v?

A

v = (100 U [10 mM]) / (40 mM + 10 mM) = 1000 U.mM / 50 mM = 20 U

58
Q

Describe and explain graphically with 1/V vs. 1/[S] plots and pictorially with drawings: • irreversible inhibition

A
  • Graphically indistinguishable from a non-competitive inhibition graph.
59
Q

Draw the Michaelis-Menten saturation curve for unhibited enzyme, competitively inhibited enzyme, non-competitively / irreversibly inhibited enzyme. Indicate what occurs to Km and Vmax as a result of inhibition when compared to uninhibited.

A