Enzymes Flashcards

1
Q

Catalyst

A

A substance that speeds up the rate of a chemical reaction but it is not consumed

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2
Q

What are some examples of catalysts?

A

most are proteins: enzymes

few are RNA- peptide bond formation is catalyzed by RNA in ribosomes

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3
Q

coenzymes

A

required by some enzymes

- needed like a sidekick for enzyme function

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4
Q

aponoenzyme/ apoprotein

A

protein component of the enzyme

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5
Q

Holoenzyme

A

Protein + coenzyme= fully functional

* not all enzymes need coenzymes

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6
Q

Why do we get nutrient deficiency?

A

you arent providing the enzyme the coenzyme

  • if its the apoenzyme alone all the time it would not be fully functional
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7
Q

oxidoreductases

A

transfer e- as H(has an e- and proton) or H+ (hydride ion, lost protons)

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8
Q

transferase

A

transfer groups between molecules

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9
Q

hydrolases

A

add functional groups to water(uses h20 to assist in cleaving covalent bonds- hydrolysis)

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10
Q

Lyases

A

form or add double bonds- cleaves covalent bonds without water often resulting in formation of new double bonds)

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11
Q

isomerases

A

isomerize by group transfer- different chemical properties due to different connections

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12
Q

Ligases

A

form C-C, C-S , C-O, C-N bonds, coupled to ATP cleavage(energy)
- covalent bond formation is entropically unfavorable ie- energetically complex

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13
Q

Why are enzymes necessary?

A
  1. Enzymes catalyze chemical reactions
    *promotes survival and allows you to get by in complex environments
    they can accelerate bond formation and breakdown by 10^6- 10^12
  2. responsible for majority of all reactions in living systems
  3. Very specific: no side reactions
    * gives control- ex- if theres a deficiency they only one reaction is affected
  4. Can be regulated
    * If a reaction happened without enzymes it would take forever
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14
Q

Explain using an example why enzymes are needed in the body?

A

People who are deficient in lactase, lactase breaks down lactose but if your deficient you don’t have it the lactose goes straight thru into the large intestine and bacteria will break it down into disaccharides

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15
Q

substrate

A

the reactant on which the enzyme acts to convert it to product
- 2 or more substrates could react together and there could be 2 or more products

ex- substrate- lactose
enzyme- lactase

glucose and galactose= products

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16
Q

Enzyme Substrate complex (ES)

A

a non covalent reversible association between enzyme and substrate

  • catalysis occurs here
  • little energy needed to disrupt it
  • s is partly distorted so it starts to look like S at TS
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17
Q

active site

A

the pocket on the enzyme where the substrate binds and the reaction is carried out

  • is like the chair you sit on and provides support
    non covalent- ionic vdw h bonds
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18
Q

Transition state

A

an activated state of increase energy which the reactants must pass thru to become products

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19
Q

What is the Transition state theory?

A

the rate of a chemical reaction depends on how much energy the reactant or substrate must acquire in order to reach TS

rate depends on temp and delta G of TS

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20
Q

What is delta g with double dagger mean?

A

the activation energy
- the extra energy that S must acquire to reach the transition state- the energy you have to put in to reach the transition state

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21
Q

Why is S double dagger of higher energy then S?

A

because S must go thru distortion ( bond stretching or bending, locking in a rare conformation in order to react

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22
Q

When S is bound to an enzyme no extra energy is needed to reach the distorted form in ES complex why?

A
  1. formation of the E.S is energy releasing (spontaneous favourable)
  2. Some of the binding energy is used to distort the substrate moving it along the reaction coordinate
  3. Even in the ground state ES complex the enzyme is already working to catalyze the reaction
23
Q

How does the activation energy get lowered?

A

the binding of the E to the S where a number of weak but favourable bonds are formed (h- bonds, hydrophobic interactions vdws)

24
Q

What are some of the effects of ES formation?

A
  1. a decrease in entropy of S(only one conformation)
    * unfavourable bc enzyme and substrate together is entropically not a benefit
  2. Desolvation- removal of H20 shell around S
  3. Induced Fit- the enzyme adjusts to the shape of transition state( this is how reactions take place enzyme is changing shape to fit transition state
  4. alignment of the groups that must react
25
Q

How does an enzyme recognize and bind a substrate?

A
  1. Shape consistency or fit- lock and key- the substrate must fit the active site- not true
  2. Electrostatic consistency- correct matching of ionic and h bonds within active site
    * vdws are not electrostatic but still important
  3. Thermodynamic consistency- can protein flex to adopt a substrate or the substrate flex to fit into the active site
26
Q

What is not effected by a catalyst ?

A

Keq
if the enzyme increases the forward rate then the back must also increase

s and P energy doesn’t change regardless of catalyzed orW
uncatalyzed

27
Q

What is binding energy use for?

A
  1. entropy reduction- hold the substrate close together in proper orientation for reaction
    * take disorder and reduce it
  2. Desolvation
    - substrate molecules are surrounded by water hydration shell that must be removed for reactions to occur
  3. Strain reduction
    - streic and or electronic strain must be accomodated
28
Q

What is enzyme specificity?

A

the enzyme is usually bigger than the substrate and the substrate binds to the active site
each enzyme catalyzes one reaction using a limited number of substrates

1) optical (chiral) specificity
- one enzyme recognizes on D AAs and another is needed to recognize L AAs

2) geometric specificity
- ex fumarase converts fumarate (trans isomer) to malate but cant use maleate(cis)- substrate recognition is very specific

29
Q

What is the catalytic triad?

A

3 AAs in the active site of chymotrypsin that are very involved in bond breaking

His 57 Asp102 Ser195

30
Q

What is the aromatic side chain pocket?

A

hydrophobic pocket (vdws interactions)
- enzyme specificity- you get vdws here
where R groups go

31
Q

Nucleophile

A

go after positivity

32
Q

electrophile

A
  • partial positive makes it subject/target to nucleophilic attack
  • attacked by things that are negative
33
Q

Why do enzymes use metal cofactors?

A
  1. Weak interactions between metals and the substrate help stabilized charge transition states and may help orient and bind substrate
  2. Metals accept and donate electrons in Redox reactions
    - allow to orient and keep substrate bound properly
    - appropriate shape to keep everything together
34
Q

Carboxypeptidase

A

uses zinc in its active site as an alternative to using AAs to form and oxyanion hole

35
Q

What does carboxypeptidase do in the body?

A

procarboxypeptidase is an enzyme in the body inactive and then carboxypeptidase activates it
it helps cut proteins into smaller pieces and cut pepsin

36
Q

What does it mean if the value of K2 is very high?

A

the the enzyme S affinity would be very low bc they want to break up

37
Q

What does it mean if K value if high for K1?

A

it means the E and S want to be together

- high affinity

38
Q

What are some assumptions made in MM?

A
  1. K3 > K4 so K4 isn’t important at start of the reaction bc at the beginning when you add substrate you don’t have product- [P=0]
  2. K1> K3 which makes E+P formation slow and the rate determining step
    * v= k3(ES)
    product formation is the rate limiting meaning the entire reaction cant go faster then it
39
Q

Whats [E total]?

A

[E total] = [E] + [ES]

[E]- free enzyme not bound to substrate

[ES]- actively engaged in reaction

40
Q

When will vmax occur?

A

when all of E is bound to S or

[Etotal}= [ES- every single enzyme needs to be actively bound to a substrate] or Vmax= k3[Etotal]

41
Q

What are the equations for Rate of formation and breakdown of E.s?

A

rate of formation of E.S: E.s= k1[E][S]

rate of breakdown of E.S: k2[E.S] + k3[E.S]

42
Q

What would you expect for K2+ K3 if enzyme likes substate?

A

K2 +K3 measure breakdown so theu be low when high affinity

K1 will be high when E likes S

43
Q

What is K m?

A

measure of affinity of enzyme for substrate

km= K2 +k3/K1

44
Q

Michaelis menten kinestics

A

linear increase in velocity and [S] increases

45
Q

What does a high/ low km mean?

A

a high Km reflects a fast dissociation or a low affinity
* need a high [S]- unfavourable reaction

a low reflects slow or limited dissociation or a high affinity
* need low [s]- favourable reaction
ideal bc not much effort needed

46
Q

What does it mean when Km= S

A

1/2 of E is bound to S or [E.S]= [E]

47
Q

What is Kcat?

A

catalytic rate constant= k3(rate determining step)

addresses how good enzyme is

if rate determining happens fast the good if it happens slow it isn’t best suited for reaction it is suited but if you had the opportunity to replace you would

48
Q

What is the specificity constant?

A

reflects the catalytic efficiency of an enzyme

  • used to compare catalytic efficiencies of different enzymes or turnover of different substrates
  • much more accurate than using one

v= Kcat/ km[Etotal][s]

49
Q

What is an enzyme inhibitor?

A

are categorized on the basis of how they effect enzyme kinetic parameters

is a compound that when added to solution with an enzyme and substrate reduces the rate of conversion S-> P

50
Q

Describe Competitive inhibitors

A

resemble the substrate and bind in the active site blocking access of the natural substances
ex- lipitor viagra

  • at high [S] the I is displaces from E so Vmax isnt changed
    km is increased bc it takes much more S to reach vmax
51
Q

Describe non-competitive inhibitors

A

I binds at a site distinct from the substrate site usually an allosteric site- I is another shape not same as the substrate

allos-other
stereos-shape

-it may bind to free E or to ES
once bound it will prevent P formation

if the binding affinities of I to E and ES are identical there will be no effect on Km
BUT since the I decreases active [Etotal] then vmax must decrease (vmax=kcat[etotal]

if affinity of I for E and ES are different “mixed inhibition” is obtained and both vmax and km are changed)-
*competitive couldnt bind to ES only free E

52
Q

Does a non competitive inhibitor affect ability of free enzyme to engage with substrate?

A

No because enzyme is still able to get together with substrate km doesn’t change

53
Q

Describe uncompetitive inhibitors

A

I binds to an allosteric site but only to the ES complex
slope of 1/v0 vs 1/[S] arent changed but vmax is lower and the [S] needed to reach 1/2 vmax= km

  • etotal is lowered
  • low vmax E and S are gonna get together to form more substrate shifiting equ’ to the right affinity is higher km is lower
54
Q

Allosteric enzymes

A

v0 vs [s] is sigmodial whereas MM plots are hyperbolic

  • enzyme is very sensitive to [s] over a narrow range and behaves like an on-off switch
  • in multisubunit enzymes
  • cooperativity( the occupancy of an active site on one subunit has an effect on the other subunits)
  • binding of S to one active site makes binding of subsequent easier