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Protein Control in Cellular Regulation > Enzymes and Ligand Binding > Flashcards

Enzymes and Ligand Binding Flashcards

1
Q

What is the reaction classified as when delta G is negative?

A

It’s a spontaneous reaction.

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2
Q

What does a spontaneous reaction mean?

A

The reaction can proceed forward because there is energy available to do work.

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3
Q

What is the speed of spontaneous reactions?

A

Slow

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4
Q

What do enzymes do to spontaneous reactions?

A

They increase the rate of the reaction by lowering the activation energy.

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5
Q

What do enzymes not influence?

A

The delta G for the reaction (difference in energy between the products and reactants).

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6
Q

How does the energy of the transition state compare to either the reactants or products?

A

It’s the most unstable state due to having the greatest amount of energy.

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7
Q

For enzyme kinetics, describe the starting rate of the reaction.

A

At the beginning, as the substrate concentration increases (still low), the rate of reaction increases exponentially.

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8
Q

For enzyme kinetics, describe the end of the rate of the reaction.

A

The rate of reaction plateaus at high substrate concentrations. No matter how much substrate is added, it doesn’t affect the reaction rate since all enzymes are occupied (1st order).

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9
Q

What order of reaction describes the plateau in enzyme kinetics?

A

1st order

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10
Q

What is the Michaelis-Menten equation?

A

v = Vmax[S] / Km + [S]

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11
Q

What is the equation for the Michaelis constant?

A

Km = (k-1 + k2) / k1

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12
Q

What does v mean in the Michaelis-Menten equation?

A

Rate of reaction

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13
Q

What does the Km mean?

A

It’s the Michaelis-Menten constant, which is the substrate concentration at half maximal velocity.

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14
Q

When Km = [S], what is the Michaelis-Menten equation?

A

v = Vmax/2

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15
Q

What are the assumptions for steady-state kinetics?

A
  1. [ES] is constant, therefore the rate of formation of ES = the rate of breakdown of ES
  2. [S]initial&raquo_space;[E]initial, therefore [S]free = [S]initial
  3. The initial concentration of P = 0, therefore back reaction can be ignored (only measure initial rates)
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16
Q

What is the equation for Vmax?

A

Vmax = [E]initial x kcat (k2)

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17
Q

What is needed for the values of the Michaelis-Menten equation for an enzyme to be efficient?

A
  1. Vmax is large
  2. Km is small
  3. k2&raquo_space;> k-1 (allowing the forward steps to proceed)
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18
Q

What is the problem with the Michaelis-Menten equation and the efficiency of an enzyme?

A

k2 cannot be both large and small. It influences both Vmax and Km directly.

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19
Q

What is the equation for catalytic efficency?

A

kcat/Km

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20
Q

How does the size of Km relate to affinity of the enzyme?

A

The lower the Km, the greater the affinity of the enzyme for its substrate.

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21
Q

How do enzymes work when physical binding to their substarte?

A

They stabilize the transition state, lowering its energy, so not a great amount of activation energy is needed for the reaction to proceed.

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22
Q

How do enzymes aid in catalysis?

A

They provide chemical groups to facilitate the reaction that allow for covalent intermediates (divides the reaction into sub-steps) and stabilizes the charges on reactants.

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23
Q

Why don’t enzyme bind tightly to substrates?

A

When bound tightly, it blocks the active site of the enzyme so they reaction can’t proceed. As it’s bound tight, it reduces the energy of the enzyme-substrate complex, reducing the rate of reaction.

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24
Q

What does the enzyme do to the transition state?

A

It tightly binds to the transition state, thus lowering its energy, so less activation energy is needed and there is an increase in the rate of reaction.

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25
Q

Why are transition-state analogues good inhibitors?

A

Due to similarities in structure to the real transition state, the enzyme binds tightly to the analogue. This inhibits the reaction as the proper active site can’t recognize the substrate.

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26
Q

Give an example of a transition-state analogue.

A

Adenylate kinase can be inhibited by Ap5A. The transition state of ATP + AMP is when one of the phosphates from ATP is stretched halfway between both substrates at the point before breakage of the bond. Ap5A looks similar to this state as it has 5 phosphates, so the enzyme (adenylate kinase) will bind tightly.

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27
Q

How to use transition state analogues for antigens?

A

Use a transition state mimic for the antibody, so that antigens can’t bind and form a complex. This inhibits the phagocytic proteins that destroy the antibody-antigen complex.

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28
Q

What do serine proteases do?

A

They catalyze the breakdown of proteins, such as trypsin or chymotrypsin.

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29
Q

What are zymogens?

A

Inactive forms of enzymes.

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30
Q

Where do serine proteases cleave proteins?

A

They cleave unfolded regions that are more exposed. These technically exist between protein domains.

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31
Q

What is the purpose of the catalytic triad of serine proteases?

A

They change the chemistry of the reaction allowing for catalysis.

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32
Q

What does the oxyanion hole of serine proteases accomplish?

A

It stabilizes the tetrahedral transition state.

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33
Q

How does a serine protease accomplish non-specific binding?

A

The substrate binding site is non-specific so any protein can bind because of the beta sheet interactions that occur. Beta sheet interactions occur between carbonyl and amino groups, which all proteins have.

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34
Q

How does the specificity pocket of a serine protease differ for chymotrypsin and trypsin?

A

Chymotrypsin = cleaves after aromatic, usually hydrophobic residues (phenylalanine, tyrosine, and tryptophan).

Trypsin = cleaves after basic, positive residues (lysine and arginine).

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35
Q

What does covalent modification do for enzymes?

A

It inactivates the enzyme through chemical modification of residues, usually through the addition of methyl groups.

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36
Q

What does X-ray crystallography determine?

A

Determines the structure of enzymes.

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37
Q

What does nuclear magnetic resonance spectroscopy (NMR) look at?

A

The ionization state of the active site of the enzyme.

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38
Q

What substrates are used for steady state experiments? Why?

A

Esters (acid and alcohol) because they can be linked to chromophores to let off colour (yellow).

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39
Q

What is measured in steady state kinetics?

A

The initial rate for different substrate concentrations.

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40
Q

What did substitution of R1 and R2 groups for chymotrypsin reveal (done through ester substrates)?

A

The R1 group affects the rate of catalysis more than the groups on R2, thus they are needed for the recoginition of the substrate for cleavage (before cleavage site). This is because they reduced the kcat value by two fold when changed.

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41
Q

What does the use of covalent modifications help to identify in an enzyme?

A

The active site residues.

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42
Q

What does DIPF react with on chymotrypsin?

A

Ser195

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43
Q

What does PMSF react with on chymotrypsin?

A

Ser195

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44
Q

What does TPCK react with on chymotrypsin?

A

His57

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45
Q

What residue is specific for chymotrypsin over trypsin?

A

Histidine

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46
Q

What happened when DIPF bound to Ser195?

A

The modification of serine revealed that it reacts to form a covalent structure and is very reactive. This indicates that it’s a good nucleophile as it donated an electron pair to DIPF.

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47
Q

What is a covalent structure?

A

Bonds between nonmetals.

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48
Q

What happened when PMSF bound to Ser195?

A

It revealed that serine reacts to form a covalent structure and is very reactive, thus making it a good nucleophile.

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49
Q

What happened when TPCK bound to His57?

A

It revealed that chymotrypsin will become inhibited because the aromatic group of histidine sits in the specificity pocket. This makes histidine 57 an important residue in the enzyme.

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50
Q

What is burst kinetics?

A

There is an initial burst (rapid increase) in reaction velocity before the steady state (the slow second step) is reached in an enzyme catalyzed reaction.

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51
Q

What did the use of p-nitrophenyl ester (substrate) for chymotrypsin reveal for burst kinetics?

A

When extrapolated at time zero, some product generated more rapidly than the steady-state rate. When the concentration of enzyme was doubled, this doubled the rate of activity and burst.

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52
Q

What does burst kinetics indicate?

A

There is an initial rapid reaction rate and a slower steady-state rate, indicating that there are two products.

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53
Q

What step in burst kinetics determines the steady-state rate?

A

The slow second step.

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54
Q

Explain the pre steady-state and steady-state kinetics of chymotrypsin.

A
  1. Substrate binds to the enzyme, forming a complex
  2. A fast initial reaction occurs, releasing a product
  3. The serine residue of the specificity pocket of chymotrypsin holds tightly
  4. The covalent structure goes through hydrolysis (addition of water); slow step
  5. Release of the second product and the enzyme can begin the cycle again
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55
Q

What residues are responsible for the reactivity of chymotrypsin?

A

Serine 195, histidine 57, and aspartate 102.

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56
Q

How is chymotrypsinogen transformed to chymotrypsin (the active form)?

A

The 245 inactive residue protein has dipeptides 14-15 and 147-148 clipped out.

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57
Q

How do the residues of the active site of chymotrypsin interact?

A
  1. Serine donates a proton to histidine
  2. This puts a positive charge on histidine and negative charge on serine, allowing it to be a nucleophile to attack the amide/ester substrate
  3. The negative charge of aspartate balances out the positive charge of histidine (there is no donation of protons between the two)
58
Q

What structure does the substrate of chymotrypsin exist in?

A

In a planar form.

59
Q

Explain the catalytic mechanism of chymotrypsin.

A
  1. Serine donates a proton to histidine
  2. The negatively charged serine (oxygen) attacks the substrate at the carbonyl
  3. The oxyanion hole stabilizes the tetrahedral structure of the transition state
  4. The first product is released and the covalent structure intermediate is released (fast step)
  5. Water is added: donates a proton to histidine and the hydroxyl group attacks the covalent intermediate
  6. An oxyanion hole stabilizes the transition state tetrahedral intermediate of the slow step
  7. The second product is released
  8. Enzyme can begin catalysis again
60
Q

Why doesn’t chymotrypsinogen work?

A

Even though it has the catalytic triad, there is no oxyanion hole, so it cannot stabilize the transition state.

61
Q

What is the purpose of site-directed mutagenesis?

A

It changes one or more amino acid residues in the protein by changing the DNA sequence. Therefore, the effect on enzymatic activity can be assayed and compared to the wild-type enzyme.

62
Q

What does each line on a log scale mean?

A

A 10 fold difference.
ex. 5 ticks between two columns means a 10^5 fold difference

63
Q

What did the substitution of alanine for serine in subtilisin reveal?

A

A 10^7 fold reduction in activity.

64
Q

What did the substitution of alanine for histidine in subtilisin reveal?

A

A 10^7 fold reduction in activity.

65
Q

What did the substitution of alanine for aspartate in subtilisin reveal?

A

A 10^5 fold reduction in activity.

66
Q

What did the substitution of alanine for all three residues in subtilisin reveal?

A

A 10^7 fold reduction in activity.

67
Q

What did the uncatalyzed form of subtilisin reveal?

A

A 10^10 fold reduction in activity from the wild-type, and a 10^4 fold reduction in activity from the fully mutated enzyme.

68
Q

Why was the mutated forms of serine and histidine more important than aspartate?

A

They are involved in the transition state and covalent intermediates, which help to form the products; however, aspartate just stabilizes the charge on histidine.

69
Q

Why does the fully mutated enzyme still have greater activity than the uncatalyzed enzyme?

A

Because the enzyme stabilizes the tetrahedral transition state through it’s oxyanion hole.

70
Q

What happens when glucose binds to hexokinase?

A

It triggers a conformational change of the enzyme that allows all the catalytic residues to be exposed at the active site where glucose sits. This allows glucose to be phosphorylated.

71
Q

How does simple binding exist on a graph?

A

It shows hyperbolic saturation as a function of ligand concentration.
ex. myoglobin binding to oxygen

72
Q

What is Kd?

A

The equilibrium dissociation constant, where the concentration of ligand is at half maximal binding.

73
Q

What is the equation for Kd?

A

Kd = [P][L]/[PL]

74
Q

What happens to the equation when [L] = Kd?

A

[P]/[PL] = 1, so [P] = [PL]

75
Q

What are units of Kd?

A

Molar (M2/M from its equation).

76
Q

What is the equation that involves delta G and Kd?

A

Delta G = -RTlnKd

77
Q

What does it mean when molecular mass is 100 Da?

A

1 mole of drug has a mass of 100 grams.

78
Q

Conversion of dalton to g/mole.

A

1 Da = 1 g/mol

79
Q

To block 90% of the binding sites, how much Kd would you need?

A

10 times its value.

80
Q

What does Kd measure?

A

The affinity of the enzyme for the substrate (tightness of binding).

81
Q

How do affinity and the size of Kd relate?

A

Smaller the Kd, the tighter the binding (higher the affinity).

82
Q

How does affinity relate to function in terms of inflammatory response?

A

The Kd value is large indicating a weak attraction between selectin and carbohydrate. The selectin exist on the endothelium of blood vessels and attaches to the carbohydrates on WBCs as they leave the vessel to fight infection. The weak interaction ensures that selectins aren’t pulled from the basement membrane.

83
Q

How does affinity relate to function in terms of antibody-antigen binding?

A

The Kd value is small indicating a tight interaction between antibodies and antigen. This makes sure that the complexes stay in tact so that phagocytic cells can come clear the antigen from the system.

84
Q

What happens when ligand concentration is in excess over protein concentration?

A

Then [L] = [L]total

85
Q

How to determine Kd from the fraction of bound protein?

A

Fraction of bound protein = [PL]/[P]total = [L]/(Kd + [L])

86
Q

What is the fraction of bound protein when [L] = Kd?

A

0.5

87
Q

What are the equilibrium methods that measure Kd?

A

Equilibrium dialysis, centrifugation, titration calorimetry, and spectrophotometric titrations.

88
Q

What are the kinetic methods that measure Kd?

A

Rapid mixing techniques like stopped-flow fluorescence and surface plasmon resonance.

89
Q

What do the kinetic methods measure to give Kd?

A

The association; reactant equation (kon) and dissociation rate constants; product equation (koff).

90
Q

What does isothermal titration calorimetry measure?

A

The heat released (if exothermic) or absorbed (if endothermic) upon binding to the cells.

91
Q

How does the temperature differ in the sample cell (buffer) and reference cell (buffer-protein)?

A

It’s the same.

92
Q

What measurements does isothermal titration calorimetry give?

A

Delta h and delta g.

93
Q

How does analytical ultracentrifugation work?

A

Molecules are spun in a centrifuge so that the distribution of their molecular weight can be measured. A complex will sediment faster due to its larger weight. The plot of the gradient can determined Kd.

94
Q

Why must happen for any method to measure Kd?

A

signal P + signal L must not equal signal PL

95
Q

Rate of association equation

A

k on x [P] x [L]

96
Q

Rate of dissociation equation

A

k off x [PL]

97
Q

What happens at equilibrium for k on and k off?

A

k on x [P] x [L] = k off x [PL]

98
Q

What equation of k on and k off give Kd?

A

Kd = k off/k on = ([P] x [L])/[PL]

99
Q

What does surface plasmon resonance do?

A

It detects changes in the refractive index upon binding in the immediate vicinity of a surface layer of a sensor chip (immobilize protein on it).

100
Q

What is a refractive index dependent on?

A

The mass of material on the surface of the chip.

101
Q

When is stopped-flow fluorescence a useful approach?

A

When:
fluorescence of protein + fluorescence of ligand does not equal fluorescence of protein-ligand complex

102
Q

How does stopped-flow fluorescence occur?

A
  1. Sample A = ligand and sample B = protein
  2. They are mixed using a pneumatic drive
  3. Detect the fluorescence of product formed to determine the Kd
103
Q

What do lysozymes do?

A

They have bactericidal activity (kill bacteria) by hydrolyzing (breaking) polysaccharide in the cell wall.

104
Q

What is peptidoglycan?

A

A network of interlinked sugars and short peptides that form the cell wall of gram-positive/negative bacteria.

105
Q

What are the sugars that make up peptidoglycan?

A

Repeating units of NAM and NAG

106
Q

How does NAG differ from NAM?

A

NAM is larger than NAG, which is important in the identification of how the lysozyme works.

107
Q

Where does lysozyme cleave NAG and NAM units?

A

On the C1 side of the glycosidic bond.

108
Q

How was it determined what side of the glycosidic bond did lysozyme cleave?

A

18O is a heavy isotope that was used as water. Using mass spectrometry, one of the cleaved monosaccharides received the oxygen.

109
Q

How many sugars fit into the active site of lysozyme?

A

Six units.

110
Q

Where does cleavage occur when the substrate is placed in the lysozyme?

A

It’s cleaved at an active site between D and E. It occurs after NAM.

111
Q

How was the cleavage site in the lysozyme determined?

A
  1. Trisaccharide of NAG placed in sites ABC = no cleavage, thus the active site doesn’t exist in these three units
  2. When NAG6 was placed, the products were NAG2 and NAG4, thus the only other applicable cleavage site is between D and E
  3. However, NAM can’t be placed in the site C as it’s too big
  4. When NAM was placed in site 4, there was hydrolysis of the protein as site D/E
112
Q

What is the optimal pH of lysosomal activity?

A

pH of 5.

113
Q

What happens to Asp52 at a low pH?

A

It’s protonated and thus no activity.

114
Q

What happens to Glu35 at a high pH?

A

It’s deprotonated and thus no activity.

115
Q

What must happen to Asp52 for catalysis for lysozymes?

A

Must be deprotonated (negative charged).

116
Q

What must happen to Glu35 for catalysis for lysozymes?

A

Must be protonated (neutral charge).

117
Q

What happened when NAG-NAG-NAG-NAG-lactone was bound to the lysozyme?

A

A very tight binding, thus a very low Kd value.

118
Q

What did the tight binding of NAG-NAG-NAG-NAG-lactone to the lysozyme suggest?

A

That NAG-NAG-NAG-NAG-lactone was the intermediate.

119
Q

Why was it wrong that NAG-NAG-NAG-NAG-lactone was an intermediate?

A

Enzymes don’t tightly bind to the intermediate, rather it’s the transition state.

120
Q

What is the most stable conformation of sugars?

A

Chair

121
Q

Why does increasing the number of NAG units lead to a lower Kd value?

A

This is because there are more interactions with the lysozyme, indicating it’s more bound to the enzyme.

122
Q

What happens to the Kd when NAM is substituted for NAG at site B?

A

Little effect as Kd value went from 8 to 3.

123
Q

What happens to the Kd when NAM is substituted for NAG at site D?

A

An increase in binding energy as Kd value went from 8 to 1.

124
Q

How does NAG3-NAG-lactone compare to NAG4 in terms of Kd value?

A

The lactone analogue has a much smaller Kd value.

125
Q

What was the carbonium ion (lactone analogue) proposed to be?

A

An intermediate.

126
Q

What was the philips mechanism of catalysis (the incorrect method)?

A
  1. Glu35 donates a proton to the oxygen in the glycosidic bond
  2. A negatively charged Glu35 now and the first product is released
  3. A carbonium ion is formed (positive carbon)
  4. Asp52, which is negatively charged for activity, stabilizes the positive charge on the planar carbonium ion intermediate
  5. Hydrolysis allows the second product to be released and the amino acids to be restored to their original ionized states
127
Q

Why could the carbonium ion and later intermediates not be detected?

A

Because the first step in the lysozyme mechanism was the slow step.

128
Q

How was the step 2 of lysozyme catalysis slowed down to allow intermediate buildup?

A

Used Glu35Gln (no negative charge) mutant of lysozyme that was much less efficient than the wild type enzyme.

129
Q

How was it proven that the carbonium ion was not the covalent intermediate?

A

When step 2 was slowed down to reveal intermediates, mass spectrometry and crystallography showed this error.

130
Q

What is the covalent intermediate formed between?

A

NAM residue and Asp52

131
Q

Why does the short existence of the carbonium ion suggest that it is a transition state?

A

Transition states are unstable due to the great amount of energy they contain, thus they don’t last long.

132
Q

Explain the Koshland mechanism (correct mechanism for lysozyme).

A
  1. Glu35 donates a proton to the oxygen in the glycosidic bond
  2. A negatively charged Glu35 now and the first product is released
  3. There is the carbonium ion formed that exists very shortly, thus a transition state
  4. The NAG product diffuses away
  5. Covalent intermediate formed between NAM residue and Asp52
  6. Hydrolysis attacks the intermediate, using the negatively charged OH group to displace Asp52
  7. The H, from H20 protonated Glu35 and both ions’ charges are restored
  8. Second product is released
133
Q

What is the pKa?

A

The pH when half the molecules are protonated and the other half are deprotonated within a population.

134
Q

How do the peptide bonds in a substrate and transition state differ?

A

In a substrate, peptide bonds are planar whereas they are tetrahedral in the transition state.

135
Q

What are units for rate of reaction?

A

M/s

136
Q

What is unit for koff?

A

s-1

137
Q

What is unit for kon?

A

M-1 x s-1

138
Q

What happens to Km if more enzyme is added?

A

It’s unchanged as it’s the intrinsic property of the enzyme. The affinity for its substrate is still the same. Km = Vmax/2, therefore enzyme concentration doesn’t affect the equation.

139
Q

What is Kcat?

A

Kcat is the turnover number. It’s the number of substrate molecule each enzyme site converts to product per unit time or kcat is the rate constant for the reaction ES –> P.

140
Q

What happens if the kcat value is high?

A

The more efficient/specific the enzyme is for its substrate.

Protein Control in Cellular Regulation (8 decks)

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