enzymes II Flashcards

1
Q

What is a perfect enzyme?

A

Perfect’ enzymes have reaction rates limited by diffusion .

evolution has improved the chemistry of the 2nd step such that the rate limiting step if the diffusion of the enzyme and substrate together

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2
Q

in an enzyme substrate reaction, which part can evolution work on?

A

, a natural process by which E and S diffuse together in solution to form the ES complex.
The second step is to do with chemistry. Evolution can work on this step, by mutations it can improve the properties of the active site so that the free energy of activation for this step is reduced, therefore evolution can improve the chemistry of the reaction.
- the overall reaction rate is determined by the slowest step.

Evolution cannot improve the physical nature of the first step, ie diffusion, as its determined by the properties of the enzyme and substrate.
In this process, the overall reaction rate is determined by the slowest step. By improving the chemistry, we will reach a point where the limiting step Is the diffusion step.

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3
Q

outline 3 experiments that can determine whether an enzyme is perfect and that diffusion is the rate limiting step.

A

If diffusion is the rate limiting step, then there overall should be a reduction in the rate of product formation.

  1. Altering the viscosity of the solution. Is the reaction rate affected by viscosity?
    Add glycerol to the reaction mixture- it makes the solution more viscous and slows-down the rate at which E and S diffuse together to form ES. If a change in viscosity slows the enzyme reaction rate then the enzyme is diffusion limited and so perfect. eg: carbonic anhydrase.
  2. A second approach to identifying whether an enzyme is perfect and limited by diffusion is by theoretical calculations.
    . if the enzyme is diffusion controlled, then the rate constant should be 108.
    Carbonic anhydrase: k3 is ~600,000 s-1 ; KM is 8 x 10-3M
  3. free energy profile. If the diffusion steps are rate limiting then the enzyme is perfect.
    - A lot of work to obtain free energies for all the reaction species and transition states but yields hard data that enables you to see whether the diffusion steps are limiting overall for the enzyme
    e. y triosephosphate isomerase
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4
Q

why is triosephosphate a good catalyst

A
  • The role is to triosephosphate is to interconvert dihydroxyacetone phosphate to glyceraldehyde 3-phosphate.
  • The result of this Is that both C3 halves of glucose can be utilized to make ATP.
  • During evolution, triosephosphate isomerase would be under intense evolutionary pressure to be a good catalyst as having a functioning and efficient glycolytic pathway would be crucial for survival because it makes ATP needed for muscle contraction
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5
Q

how can we tell that triosephosphate is a perfect enzyme?

A

the free energy profile

  • The enzyme and substrate forms the ES complex – diffusion step.
  • The ES complex is converted into an enzyme intermediate complex EZ – chemical step
  • EZ is converted into the enzyme bound product – chemical step.
  • The product is released – diffusion step.
  • In the free energy the rate limiting steps are the diffusion steps because evolution has reduced the chemical steps to the point where they are less efficient than the diffusion steps.
  • There is no point in the chemistry reducing the free energies of activation of these steps any further because it doesn’t affect the overall rate of the reaction.
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6
Q

serine proteases are an example of?

A

reducing the free energy of activation by offering a different reaction pathway

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7
Q

describe serine proteases and how they react

A
  • Hydrolyse peptide bonds
  • The enzymes take place in the chemical steps involved in the hydrolysis of the peptide bonds . it forms an acyl enzyme intermediate with the substrate.
    ]- 3 serine proteases: Chymotrypsin and trypsin, both secreted by the pancreas and involved in the digestion and elastase
  • All three of these enzymes, and other serine proteases have an active serine at their active sites that attacks the peptide bond to form an acyl enzyme
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8
Q

describe the structure of chymotrypsin

A

serine-195 is present in a groove that binds the polypeptide chain that is going to be hydrolysed

  • The active serine-195 is present in the binding pocket of the active site cleft . it participates directly in the catalytic reaction. It takes part in the chemistry of the hydrolysis of the peptide bond.
  • It is close to two other residues, histidine-57 and aspartly-102. The hydrogen bonded network between the sidechain of these residues are responsible for potentiating the reactivity of serine. catalytic triad – feature of all serine proteases.
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9
Q

how to serine proteases hydrolyse peptide bonds?

A

Serine proteases hydrolyses protein peptide bonds with sequence specificity. They don’t hydrolyse every peptide bond

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10
Q

why do serine proteases hydrolyse with sequence specificity?

A

it depends on the nature of the residue on the n terminal side of the peptide bond that’s going to be hydrolysed. Chymotrypsin prefers to cleave this bond when the residue is hydrophobic (ph,try,tyro)

  • Typsin cleaves the bond when the residue is positively charged, lysine or arginine
  • Elastase cleaves the bond when the residue is small .
  • X ray crystallography explains why :
  • Chymotrypsin has a hydrophobic pocket with asp at the bottom which accommodates the residues which are also hydrophobic. The binding pocket of trypsin has a negative charge which can form an electrostatic attraction when the residues are positive. In elastase, the binding pocket is narrow due to the valine residues
  • They show selectivity due to the nature of the pockets which bind to the reside.
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11
Q

why are serine proteases effective at hydrolysing peptide bonds ?

A
  1. They use a different reaction pathway to hydrolyse the bond compared to the uncatalyzed reaction
  2. To do this they use the highly reactive serine-OH group in the active site of all serine proteases, which catalyses the breakage of the peptide bond by a 2 step mechanism.
  3. Step 1, acylation: Instead of attacking with water (uncatalyzed reaction) uses serine-OH and attacks the carbonyl group of the peptide bond.
  4. Forms the tetrahedral intermediate which collapses. An acyl enzyme intermediate which has an ester bond forms. The amino terminal end of the pb is released.
  5. Step 2 : water attacks the acyl enzyme intermediate. The water attacks the ketone group to form a tetrahedral intermediate which collapses. This releases the enzyme which returns to its original state. In the process, it releases the carboxyl terminal end of the cleaved peptide bond.
  6. The attack of water on the acyl enzyme intermediate is easier than the attack of water on an amide group, thus this different reaction pathway hydrolyses the peptide bond using 2 steps with lower free energies of activation than that of the uncatalyzed reaction.
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12
Q

why is serine OH more reactive than OH

A

In the active site, serine-oh forms a hydrogen bonded network with the side chains of histidine and aspartate. This hydrogen bonded network forms a charge relay system which tends to pull the proton off serine-OH and onto the imidazole group and then the proton on the imidazole group is shifted onto the aspartate. the proton is being shifted away from the oxygen of serine which makes the oxygen very electronegative, therefore its able to carry out its nucleophilic attack on the peptide bond.
- The charge relay system formed by the catalytic triad of amino acids causes the effective attack of the Serine-Oh from the enzyme

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13
Q

what forms the catalytic triad?

A

serine-195, histidine-57, aspartyl-102

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14
Q

what is conserved among members of serine proteases ?

A

the 3D structure and the cleft for protein hydrolysis is conserved among members of the serine proteases.
-charge relay system

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15
Q

serine proteases hydrolyse peptide bonds through which intermediate ?

A

acyl-enzyme intermediate

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16
Q

what is the mitochondiral ATP synthase ?

A
  • Synthesises ATP- the last step in oxidative phosphorylation  this enzyme is a rotary machine driven by protons, and this rotation results in ATP synthesis.

On the inner mitochondrial membrane, facing inward, is the ATP synthase  comprises a stalk which fits into the membrane and a globular head which is in the lumen of the mitochondria.

17
Q

what enables ATP synthase to make ATP ?

A

proton gradient

18
Q

describe the structure of ATP synthase

A

ATP synthase: a multisubunit complex composed of two components. -stator it has 6 subunits with a central core, which is held in place by proteins which are anchored in the membrane. There is a rotating spindle which fits inside the hole in the stator. The spindle is in contact with 3 active sites at which the ATP is made. In the presence of the proton motive force, the proton gradient, the protons rotate the rotor. As the rotor rotates it brings in to register the three active sites inside the stator that are used to make ATP.

  • The central spindle, which is rotating, contacts three active sites
  • So proton gradient is used to drive the synthesis of ATP by rotating the machine so ADP and phosphate can be made to form ATP at 3 active sites.
19
Q

how do bacteria synthesise ATP ?

A

Bacterial also have a rotatory ATP sythase.

  • Bacteria use a proton gradient to synthesis ATP
  • The antituberculosis drug bedaquiline kills mycobacterium tuberculosis by inhibiting its ATP synthase.
20
Q

describe the enzyme TOPO II and its function

A

Top II: uses ATP to act as an enzyme clamp that unlinks daughter chromosomes.

  • Relevant for mitosis
  • The chromosomes which has been replicated and condensed In mitosis. If there is any tangling during this process, then the pulling of the microtubules to the mitotic poles could result in breakage of the DNA.
  • Topo II is able to cross one DNA segment through another and is able to mediate the untangling of chromosomes to allow their proper segregation at mitosis.
21
Q

How does topoisomerase II cross a DNA segment ?

A

Topoisomerase II and other type II enzymes cross a DNA segment from one chromosome through a transient ds-DNA break in another. -uses atp

  • Topo II is bound to a segment of the chromosome, called the G-segment.
  • It uses atp to capture a segment from another chromosome, T segment, passes it through a double strand break in the first chromosome and out through the other side of the enzyme
  • ATP resets the system for another cycle.
  • It’s a molecular machine that uses 3 gates to untangle chromosomes prior to their segregation at mitosis
22
Q

describe the steps in ATP catalysis

A
  • begins when protons pass through the part of the enzyme ATP synthase that lies in the cell membrane causing it to turn

the central core then rotates inside the top half of the enzyme

This region holds an ATP molecule and pulls in ADP and an inorganic phosphate group .

As the core rotates, the subunit with ATP loosens and the section holding ADP closes

the original ATP molecule is released and a new one is formed from ADP.

the cycle repeats.

23
Q

what would adding large amounts of substrates mean ?

A

all enzymes are bound as an ES complex and you would measure the turnover and Vmax of the enzyme

24
Q

How can the rotation of ATP synthase be visualised ?

A
  • Assembled relevant components of mitochondria by attaching them to a microscope slide. : a stator and the spindle of the rotor present in the complex. An actin filament was attached at its end to the rotating rotor. Advantage of actin is the filaments can be seen under a light microscope if you bind a ligand called phalloidin, a fluorescent ligand. – you cant see whats happening on the motor but the consequences on the actin filament of rotation by the rotor.
  • Enzymes work in both directions – atp was added, the actin filaments rotated – shows that atp synthase is an enzyme that rotates and therefore will rotate in the opposite direction when its synthesising atp .
25
Q

when an enzyme and substrate molecule come together to form an ES complex, what can happen? and what to the rates of this step depend on?

A
  • this complex can either disassociate back to the enzyme and the substrate or it can undergo a chemical reaction to form and release the product.
  • the rates of these individual steps in this process are dependent on intrinsic rate constants : K1,K2,K3.
26
Q

what does K1, K2,K3 measure

A

the propensity for a particular molecule to undergo that process.

27
Q

where does the ES scheme apply to ?

A

the rate at which the product is formed at the initial part of the reaction, the level of product is low, so we can ignore the back reaction of the product binding the enzyme.

so the es scheme applies to the reaction velocities that we measure initially at the different substrate reactions

28
Q

what is K3

A

when all the enzymes present as an ES substrate.

29
Q

what are nanomachines and give examples

A

enzymes that convert the energy stored in ATP into mechanical motion or vice versa. e.g rotary ATP synthase of mitochondria and DNA topoisomerase II