Lecture 9 & 10, enzymatics Flashcards

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1
Q

What is a common catalytic domain in enzymes (specifically serine proteases)?

  • Describe:
  • Convergent evolution?
A
Catalytic triad proteases
S H D
Or
D H S
- Describe:
OH from Ser interacts with one N from His, O- on Asp interacts with the N - H on His.
Stabilizes the pocket.
- Convergent evolution?
This has evolved in multiple different protein backgrounds. Many enzymes have different fold converging into the same catalytic domains.
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2
Q

Kinetics involves consideration of what three things?

A

Distance (bond distance for example)
Energy
Time (such as rates)

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3
Q

Keq for A + B -> C + D?

- Units?

A
keq = ([C] [D} / [A] [B])
keq = k(forward) / K(reverse)
- Units?
Depends. For
2A + B -> C + D    K units of M ^ -1
A + B -> 2C + 2D    K units of M ^ 2
etc.
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4
Q

Arrhenius Equation:

A

K = A * e ^ (-ΔG# / (R * T))
K = forward or reverse reaction rate
A = constant which describes how often molecules collide in the correct orientation when all reactants are at 1 mol per liter (1 molar)
e is e in math
ΔG# = the energy of activation
R = universal gas constant, 8.314 J / mol. K
T = temp in kelvin

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5
Q

Are enzymes complementary for the substrate, the transition state, or the product?
Why?

A

Transition state

  • Why?
    1: If the substrate was stabilized, its free energy is reduced (DeltaG), it must over come an even larger activation energy then to become the product.
    2: Product wouldn’t effect the reaction, though it might stop the enzyme from unbinding.
    3: Being complementary to the transition state lowers the deltaG during the period when its highest (the activation energy), this is what needed to become smaller
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6
Q

ΔG = ?

A

ΔH - TΔS

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7
Q

desolvation:

A

the process of dissociating or releasing water molecules electrostatistically bound to a particle in aqueous solution.

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8
Q

Ground state destabilization:

  • Define it:
  • How is it counterbalanced?
A
  • Define it:
    Loss of entropy due to forming specific enzyme substrate complex. This means you have made the substrate less stable. Since it is less stable, it will cost less energy to stabilize it (you’ve essentially moved it closer towards the top of the activation energy cost, but in a single small step)
  • How is it counterbalanced?
    Desolvation when the substrate adheres frees water, which lowers entropy cost to water. May counter balance the decrease in entropy caused by substrate enzyme complex formation.
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9
Q

When you form an enzyme substrate complex, you often increase electrostatic interaction. How?

A

You have moved the interaction into the more hydrophobic protein, lowered the dielectric, increased the electrostatic interaction strength.

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10
Q

Near attack geometry:

- Other name:

A

Theory that enzymes work by moving a nucleophile near to the electrophile. By holding them near each other, and giving them the right orientation, you increase the odds of the reaction occurring.
- Other name:
Catalysis by approximation

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11
Q

Enthalpy entropy balance in protein surface:

A

The surface of protein has more loose loops, which allows them to have many interactions with water, and have a high entropy (they can move wherever).
This entropy of the surface helps to counteract the entropy loss of binding the substrate.
While the surface is soft and moves, interacting with water the core is rigid and more constant.

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12
Q

Name five fundamentals of enzyme catalysis:

A

1: Transition state complementarity (explains why the enzyme must be complementary to the substrates transition state)
2: Ground State Destabilization (loss of entropy when substrate binds, destabilizes it, makes it easier to react into the product state)
3: Increased electrostatics interaction due to desolvation (you’ve placed ions in a more hydrophobic pocket, they bond more strongly in the catalytic site)
4: Near attack conformation (groups which need to attack each other are positioned near each other in enzyme substrate complex)
5: Induced fit (the enzyme is designed to be more relaxed when it is placing the substrate in the ideal place to react, allowing the enzyme to relax lowers activation energy cost.

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13
Q

General Acid/Base catalysis:

A

Catalysis in which a molecule (other than water) acts as a proton donor or acceptor in the reaction. Initial protonation states are formed in the end.
- Essentially, they help to stabilize the charge as it is moved.

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14
Q

Covalent catalysis:

A

Enzyme substrate complex formed with hydroxyl or thiol side chains (Cys or Ser commonly)

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15
Q

Oxidoreductase:

  • Define:
  • Other names:
A
  • Define:
    Shifting of electrons around (something is oxidized, something else is reduced)
  • Other names:
    Dehydrogenase, oxidase, monooxygenyase, hydroxylase, reductase
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16
Q

Transferase:

  • Define:
  • Other names:
A
  • Define:
    Mediate transfer of a group from one enzyme to another (like a phosphate, methyl or ubiquitin)
  • Other names:
    kinase, transaminase, transmethylase
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17
Q

Hydrolases:

  • Define:
  • Other names:
A
  • Define:
    mediate breaking of bonds by addition of water, (or formation of bonds by removal of water, but not as common)
  • Other names:
    phosphatase, peptidase, esterase, lipase
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18
Q

What system uses a lot of redox reactions?

A

Metabolism

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19
Q

Why is NADH useful for studies in enzymes?

A

It has resonance rings, and when it becomes NAD+ it gains resonance, which means you can detect this difference in UV.
NADH is detected at higher UV wavelength.

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20
Q

What are two other groups with different UV spectrum absorptions depending on if they are oxidized or reduced?

A

Flavin prosthetic group

Iron porphyrin

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21
Q

How can you tell the oxidation state of iron by UV light?

A

Fe2+ has more electron then Fe3+, will result in it being larger, pushes the perforin to expand, gives it a slightly different UV absorption pattern

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22
Q

Substrate reactivity increasing with delocalization of _______.
How does this relate to chemical groups?

A

Electrons
Also called electron dispersion
thiolester > Ester > Amide

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23
Q

Draw an ester reacting with water in hydrolysis.

A

YAY!

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24
Q

Reactions with chromophoric substrate:

A

A reaction product is generated which is detectable by UV radiation

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25
Q

You can attack serine to tag it with a chromophoric substrate.
What reacts with what?
How would you figure out what molecule you have?

A
  • What reacts with what?
    OH of serine attacks the sulfur. Proton is lost from serine. The F makes like a leaving group and leaves.
  • How would you figure out what molecule you have?
    Run it through mass spec, figure out what is attached to the tag.
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26
Q

If you continue to run enzymes in an in vitro system, what usually happens to them?

A

The enzyme eventually starts to slow, it loses its cofactor, its catalysis is not as smooth once it has turned over too many times.

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27
Q

If enzymes degrade when you run them too much how do you get a curve of velocity vs substrate conc. ?

A

You measure initial reaction rate with [S] > > [E], and only measure for a short time. Then you repeat this with less and less enzyme, and use all these points to make a curve (likely logarithmic or sigmoidal)

  • Have to keep pH constant
  • Have to keep temperature constant
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28
Q

Is E + S - > ES or ES -> E + P the rate limiting step?

A

ES -> E + P

So this is kcat

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29
Q

V = Vmax [S] / ([S]+ Km)

A

V = Vmax [S] / ([S]+ Km)

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30
Q

Km = ?

A

Km = Breakdown rate / formation rate
Km = (k-1 + kcat) / k1
Km is in units of Molarity (the conc. of the substrate will be

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31
Q

What is Km?

A

The concentration of substrate at which the enzyme will run at half its maximal rate.

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32
Q

What does Km tell you?

A

With how much affinity the enzyme binds and catalyzes the substrate. A low Km means it is very specific for that site.

High Km suggests weak substrate binding.

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33
Q

Catalytic triad:

A

Many catalytic sites will have a Acid, Base and Nucleophile (I.E. Asp, His, Ser).
They will also have an oxyanion hole. (but this isn’t part of the triad
Acid: proton donor (I.E. Asp, Glu, His)
Base: Proton acceptor (I.E. His, Ser, lysine)
Nucleophile: often will be deprotonated by (I.E.

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34
Q

How will cleavage by catalytic triad normally work?

A

Asp stabilizes a H on His, His takes the H from serine. Ser, now O- makes an attack, usually at an ester, thiolester and sometimes at an amide

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35
Q

oxyanion hole:

A

Stabilizes the anionic intermediate. Usually involves a basic amino acid, an amide or the NH from the backbones of Ser and perhaps a glycine.

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36
Q

Kcat units:

A

s^-1

are inverse time. The number of reactions which will happen in 1 second

37
Q

P-nitrophenylate:

A

Molecules can have an ester bond which is going to yield P-nitrophenylate. P-nitrophenylate is florescent at 410 wavelength.
- Allows you to track the rate of the reaction. (by seeing product formation)

38
Q

Methods of finding the active catalytic site?

A

Diisopropylphosphoflouridate

Aminoethylbenzenesulfonyl floride

39
Q

What is rate equal to?

A

The conc. of reactant * rate constant.

E.G. Kcat * [ES] = rate of E + P formation

40
Q

Assumptions of the Michaelis menton equation/model:

A

Substrate is much greater than the enzyme in concentration

That the [P] is in very small concentration and therefore K-2 is negligible.

41
Q

Michaelis menton constant:

A

Km = (K-1 + Kcat) / k1

Breakdown of ES / formation of enzyme substrate complex

42
Q

V = kcat[ES]
(as kcat is the rate limiting step)
When can Vmax be calculated?

A

Vmax = kcat[Et)

Assuming all enzyme is in ES form. (aka the [S] is ridiculously high)

43
Q

How can you check to see if you are writing the michaelis menten equation correctly?

A

If a put it the same [S] as Km I will get 1/2 the number used for vMax

44
Q

How is catalytic efficiency measured?

A

Kcat / Km
1/s / 1/M
1/s * 1/M

45
Q

In the reaction he gave p-nitrophenylate is our visible reporter, but it becomes visible before the rate the limiting step. How can this be measured?

A

Pre-steady state kinetics (measured by a stopped-flow machine)
A machine uses plungers to quickly propel the substrate and enzyme into the mixing chamber, and then to a fluorescence reading chamber

46
Q

How fast of a velocity can stopped flow methods measure?

- Why the limit?

A

In the milliseconds
Because if you push the fluid to fast in stopped flow methods then you end up with laminar flow and the fluids don’t completely mix in the mixing chamber.

47
Q

What will stopped flow methods yield?

A

A larger slope, which eventually becomes a smaller slope. The larger slope is called the burst phase, and corresponds with the part of the reaction which first yields your florescent marker.
- In other words, the first line represents binding to enzyme, and initial reactions, the second line represents a Kcat measurement

48
Q

What is the quick phase in serine a protease?

What is the slow phase?

A

Acylation is quick (can only be seen by stopped flow methods, can be inferred experimentally)
Deacylation is the rate limiting step.

49
Q

What are three different examples of serine proteases and their specificity?

A
Chymotrypsin: large hydrophobic
- has large entry, to hydrophobic domain
Trypsin: Specific for K and R.
- Asp at the bottom of the enzyme pocket
Elastase: specific for small residues
- Has two V which restrict entry to the catalytic triad
50
Q

I see a burk lineweaver plot with parallel lines what does that mean?

A

Indicative of covalent catalysis (ping pong mechanism) in a bimolecular reaction.
One molecule goes in, and is attached to the enzyme, the next molecule enters, is modified, and the reaction returns to normal.

51
Q

What is an example of the ping pong mechanism?

A

Erythrocytic nucleoside diphosphokinase II

52
Q

What is a non-sequential bimolecular enzymatic mechanism like?

A

The enzyme binds product A first, product A modifies the enzyme, the modified enzyme can bind product B, product, and catalyze its reaction. This will return it to normal.
- Slope lineweaver burk will be independent of substrate 2 conc.

53
Q

What is a sequential bimolecular enzymatic mechanism like?

A

There are two.
Ordered: means that either substrates A or B must enter first, then the other can.
Random: The substrates can enter in either order.
No modified intermediate.
- Lines will have varying slopes

54
Q

Burst phase:

A

The non rate determining step. It is acylation in a serine phosphatase

55
Q

If I have kinetics which suggest a sequential bimolecular reaction, what probably happens with my to substrates?

A

One will attack the other (since no covalent bond is made in this reaction)

56
Q

What is magnesium often used for?

A

Stabilize negatively charged intermediates, such as phosphate intermediates.

57
Q

What’s an example of a protein which uses the random mechanism?

A

Creatine kinase

Myokinase

58
Q

Substrate inhibition:

- How will this look on a Michaelis Menton plot?

A

When the substrate inhibits the enzyme which catalyzes it.
- How will this look on a Michaelis Menton plot?
Velocity of the enzyme will rise as [S] increases at first, like normal, it will then peak and go back down.

59
Q

What is a heterotrophic allosteric modulator?

A

It is a molecule other than the substrate which binds and activates or deactivates an enzyme

60
Q

ATCase:

  • General function:
  • Regulation?
  • Structure:
A
  • General function:
    ATCase is a precursor to CTP production.
  • Regulation?
    1:
    One site binds CTP, pyrimidine, and is an inhibitor
    Another site binds ATP and is an inhibitor.
    2:
    Substrate also promotes binding by modifying the formation of the other subunits.
  • Structure:
    2 C(3) units (Enzyme has 2 catalytic trimers)
    3 R(2) units (Enzyme has 3 regulatory dimers)
    Catalytic trimers rest on each other like sandwhiches, with the regulatory dimers attaching to each corner of the triangle, and contacting both Catalytic trimers.
61
Q

What is interesting about the catalytic site of ATCase?

A

There is a lysine and arginine which contribute to the catalytic site but that are not from the same subunit. In other words each subunit contributes to each of the other subunits.

62
Q

What two forms can an enzyme assume?

A

A T state and an R state
T = Tense
R = Relaxed

63
Q

When will ATCase be in the R state, when will ATCase be in the T state?

A

Binding of CTP will place ATCase in the tense state. Meaning that the enzyme is being inhibited.
Binding of the substrate (aspartate and carbamoyl phosphate) will induce a relaxing conformational change. Increase catalytic activity.

64
Q

Bisubstrate analog:

A

An analog of the transition state, but modified to be less likely to relax. For example it could be exactly the same chemically as the transition state but the leaving group has a C instead of an O at the site it would leave… so it will not leave.

65
Q

What is the purpose of a bisubstrate analog?

A

It can stay in the catalytic site. This can allow you to see how the proteins amino acids interact with the catalytic site in a crystal structure.

66
Q

What is a sigmoidal shape of a michealis mentin curve indicative of?

A

Allosteric activation.
The enzyme starts in the T-state. Binding of the substrate is disfavored.
Initial slow binding.
Binding transition to the R-state (because that is how allosteric activation do)
This transition increases the amount of product formed. Rapidly climbing V in response to [S]
Eventually all enzyme is converted to R-state, and then v max is reached. it levels off.

67
Q

What will happen to michealis menten plot for ATCase in the presence of CTP? ATP?
What is its shape normally?

A

CTP: shifted right, Km raised, Vmax lowered?
ATP: shifted left, Km lowered, Vmax raised?
Sigmoid

68
Q

If a bisubstrate analogue is good, what will be low?

A

The Kd (dissociation constant), if it is low that means it is binding with high affinity in the pocket. Wallah your bisubtrate analog is fire!

69
Q

For myoglobin or hemoglobin, Kd will be used. Why Kd?

A

Because they just associate and disassociate. Their cannot be a Kf.
Still these look like michaelis menten kinetics.

70
Q

Homotrophic allostery

A

Substrate causes the change. Examples:
Hemoglobin
ATCase

71
Q

How does the addition of oxygen change hemoglobins structure?

A

Normally iron is in the Fe2+ conformation.
At this size, it is too large to fit inside the heme perfectly, it is slightly outside the frame, being held into the frame by the proximal histidine.
When oxygen binds to Fe2+ in the heme, it takes an electron from the heme, forming Fe3+ hemeglobin. The Fe3+ is now thin enough that it can fit inside the heme ring, it slides into the heme ring (about .5 angstroms up. This allows the histidine to move up towards the heme as well. This in turn allows the alpha helix the proximal histidine is attached to to move up towards the heme. This relaxes the protein structure, as well as the structure of nearby subunits.

72
Q

How specifically is the alpha helix of the proximal histidine effected to induce the R-state when iron is oxidized?

A

It can rotate about 15 degrees, which give the enzyme more wiggle room.

73
Q

How does pH effect hemoglobin?

A

You refer to the bohr effect.
When it is lower pH, surface histidines can get protonate. When protonated they take up more space. This causes the protein to be pushed into the T state formation, and for oxygen to be more likely to disassociate.
The pH will be lower in the tissue, higher in the lung. This works for us in maximizing O2 absorption and delivery.

74
Q

What other than the bohr effect promotes release of oxygen in the tissues?

A

Heterotrophic allosteric regulation by the binding of CO2 to the N termini of hemoglobin subunits. This induces a T state and causes additional release of O2.

75
Q

What is found in the red blood cells themselves which regulates oxygen affinity?

A

2, 3 diphosphoglycerate (2, 3 DPG)

76
Q

2, 3 diphosphoglycerate binding site?

A
Interact with the beta1 and beta2 subunit, near its N terminus. This results in an increase in the amount of R form.
Beta 1:
His, Lys, His
Beta 2:
His, Lys, His
Symmetrical, with Lys across from Lys
77
Q

What is the charge of 2, 3 DPG?

A

It is pretty negative -5 across the molecule (2 phosphates, both -2, and a CO2-

78
Q

How does 2, 3 diphosphoglycerate help with high altitude?

A

It can be upregulated, decreases overall oxygen absorption, but allows a greater portion to be place in your cells?

79
Q

How does fetal hemoglobin differ from fetal hemoglobin?

A

Fetal hemoglobin has a modified site for the binding of 2,3 biphosphoglycerate (2,3 DBP). Instead of His, Lys, His it is Ser, Lys, His. Which decreases how much interaction it will have with 2, 3 DBP (which is expressed at the same level).
This allows oxygen to flow from maternal oxygen, which releases it easier (maternal more T-state), to the fetal oxygen, which holds it better (more R state)

80
Q

For a ligand and macromolecule binding and unbinding what are the equations?
A + B -> AB

A
A + B -> AB
formula is products over reactants.
Association = 
Keq = Ka = [AB] / [A] * [B]
Keq * [A] * [B] = association rate
Dissassociation = 
Kd = [A] * [B] / [AB]
Kd * [AB] = disassociation rate.
81
Q

What is true of association / disassociation at equilibrium?

A

Kd * [AB] = Ka * [A] * [B]

82
Q

Give the equation for fractional saturation:

- What are its assumptions?

A

Y = [Lt] / [RLt] / (RLt + [R])
Y = How much of the enzyme is associated with the macromolecule
[RL]/[Rt] = [Lt] / (Kd + [Lt])
[RL] / [Rt] = percent saturation (how much of the macromolecule has complexed with the ligand0
This corresponds to affinity.
- What are its assumptions?
That [Lt]&raquo_space; [Rt]

83
Q

How can you measure the amount of ligands associated / disassociated?

A

You can do a ligand dialysis screening.

Ideally you use UV or other wavelength dependent methods.

84
Q

Cooperativity:

A

Positive: binding increases affinity, slope of the hill factor > 1
no cooperativity: binding of ligand doesn’t effect affinity, slope/hill factor = 1
negative: binding decreases affinity (like substrate inhibition), slope/hill factor < 1

85
Q

Hill plot:

A
Log(Y/(1 - Y)) = nh * Log[L] - Log[Kd]
nh = hill factor = the slope
(Y/(1 - Y)) = The amount of macromolecule saturated, divided the amount not saturated
L = ligand
Kd = log of disassociation constant.
86
Q

How can stopped flow be used to get Kon and Koff rates for ligand-molecule binding complex?

A

Take my ligand, add it to excess macromolecule, measure the percentage of macromolecule which is bound using stopped flow kinetics.
Plot percent enzyme bound against

87
Q

Kd = ?

A

Kd = koff / kon

Dissassociating, koff is more important

88
Q

What does the y intersect = ? in stopped flow?

A

Koff (units of inverse seconds)
The Y intercept is when it goes off the graph
Koff

89
Q

What does the slope equal in stopped flow experiment

A

Kon (units of inverse seconds and inverse molarity)
The slope is on the graph
Kon