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Biosci 106 > 02 Catalysing Life > Flashcards

02 Catalysing Life Flashcards

1
Q

What is the function of an enzyme?

A

speed up reactions by lowering the activation energy of a reaction but do not produce any more product that a non catalysed reaction

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2
Q

Define the term Gibbs free energy

A

Free energy is the energy in a physical system that can be converted into work at a uniform temperature and pressure

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3
Q

When is Gibbs free energy negative? Why is this important?

A

When the overall entropy of the universe is increased Gibbs free energy is negative. this is important because Gibbs must be negative for a reaction to occur spontaneously (No energy input)

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4
Q

What does ΔG‡ refer to ?

A

ΔG‡ is the activation energy required to initiate a reaction

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5
Q

What is the difference between ΔG and ΔG‡?

A

ΔG is the overall free energy change in a reaction

ΔG‡ is the activation energy required to initiate a reaction

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6
Q

How do enzymes relate to ΔG‡?

A

Enzymes reduce the activation energy ΔG‡ of the reaction to increase reaction rate

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7
Q

How doe enzymes reduce the activation energy of the reaction?

A

they form an enzyme-substrate complex

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8
Q

What are the three ways a substrate may bind to an enzyme?

A

Induced fit, lock and key, conformational selection

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9
Q

What is meant by the term bonding energy?

A

The free energy that is released by the formation of weak bonds between substrate and enzyme. It drives the lowering of the activation energy in catalysis

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10
Q

When is binding energy maximised?

A

When the substrate is in the transition state.

The transition state represents the tightest interaction between substrate and enzyme. Transition state is also the least stable chemical form of the substrate

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11
Q

What is the significance of the transition state in terms of enzymatic activity?

A

the transition state can be seen as the moment at which the bond decided if it will break or reform

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12
Q

What is the function of amino acids in enzymatic activity?

A
  1. the active site of an enzyme where catalysis occurs only involves a small number of amino acids

the rest of the amino acids are for:

  1. positioning the active site residues correctly to drive catalysis
  2. providing the correct micro-environment in the active site eg shifting of pKa values (all of the proteins are involved in holding aa in certain conformation)
  3. provide other sites for recognition and control e.g. allostery
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13
Q

How are enzymes regulated?

A
  1. proenzyme/ proteolysis - one way on switch e.g. digestive enzymes
  2. proteolytic breakdown - one way off switch (breakdown when no longer needed)
  3. transient covalent modification (two way on/off switch) e.g phosphorylation
  4. Allostery (dimmer switch, graded response to non substrate molecules)
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14
Q

What are co-factors?

A

a non-protein chemical compound or metallic ion that is required for an enzyme’s role as a catalyst

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15
Q

What is the name for organic co-factors?

A

co-enzymes (derived from vitamins)

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16
Q

What are prosthetic groups?

A

Tightly bound co-enzymes that do not dissociate from their enzyme and regulate their on/off function

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17
Q

How do enzymes change the activation energy?

A

molecular recognition - enzyme substrate complex

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18
Q

How does binding change activation energy?

A

formation of weak non-covalent bonds

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19
Q

When does an enzyme recognise a substrate?

A
  • optimally recognises halfway point of the reaction. i.e. transition state
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20
Q

Where does the chemistry happen in a reaction with an enzyme?

A

the active site of the enzyme

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21
Q

What is an enzyme without its co-factor called?

A

apo-enzyme

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22
Q

What is the name of the fully functional enzyme with its co-factor bound?

A

holo-enzyme

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23
Q

What is specificity?

A

exquisite recognition of substrate by enzyme

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24
Q

What is promiscuity?

A

when enzymes show a broad range of enzyme catalysis. catalyses a transformation of a range of similar substrates

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25
Q

Why is promiscuity (“moonlighting”) a key to evolutionary advantage?

A
  • key to obtaining redundancy, resilience and adaptability in biological systems
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26
Q

What does ka and kd refer to?

A

ka is the rate of binding event i.e. association

kd is the rate of dissociation

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27
Q

what does Kd express?

A

The Kd of an enzymatic reaction expresses the ligand-receptor affinity.

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28
Q

What is the equation for Kd?

A

Kd = [P][L] / [PL] or kd / ka

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29
Q

What occurs when [L] = Kd?

A

at equilibrium, when [L] = Kd, the amount of free protein ([P]) and the amount of ligand-bound protein ([PL]) are equal.

30
Q

What does a small Kd indicate? Explain.

A

High affinity. The smaller amount of ligand you need in solution to half saturate the protein, the tighter they bind

31
Q

What does the fraction of occupancy at 0.5 tell us about an enzyme and substrate?

A

The lower the substrate concentration is at 0.5 of protein occupancy, the tighter the protein and substrate binds as less substrate is needed to half saturate the protein. I.e higher affinity, low kd, binds more tightly

32
Q

What is the the equation of the two step process Michaelis and Menten came up with for enzyme catalysis? Explain what each step indicates.

A

E+S ⇋ (ka, kd) ES ⇋ (kcat) → E+P

step 1. enzyme binds to substrate (molecular recognition, lower activation energy) (example of protein binding to ligand, ka and kd of association and dissociation, on off rate)

step 2. enzyme substrate complex turns to enzyme and product , catalytic process. Kcat is rate at which enzyme substrate complex breaks down into Enzyme + product ( one way trip ,over energy barrier)

33
Q

What is the Michaelis Menten equation?

A

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

34
Q

What is Vmax?

A

maximum rate the enzyme can go with a saturating amount of substrate

35
Q

What is KM

A

the Michaelis constant and describes the substrate concentration at which the reaction velocity is 50% of the Vmax

36
Q

What are the 4 key assumptions that must be met to use the Michaelis equation?

A
  1. catalysis is the slowest (rate limiting) step (second step of equation is the slowest ES → E +P) i.e. ka > kcat (binding is quick, catalysis is slow)
  2. *here is much more substrate than enzyme [S]≫[E]
  3. The concentration of the enzyme-substrate complex is constant: the “steady state” assumption. You end with the same concentration of enzyme-substrate complex you started with (MAJOR LIMITING ASSUMPTION!)
  4. The reverse reaction is negligible (enzyme + product does not go to enzyme substrate (this is the ball rolling back up the hill, energy) there is no reverse reaction
37
Q

What are inhibitors?

A

Substances that slow or stop enzyme reaction

38
Q

In general terms, how do inhibitors work?

A

they stop enzyme reactions by binding to the enzyme but are not chemically altered by the enzyme

39
Q

What are the tree main types of inhibition? What are the two classes of inhibition these can fall under?

A

competitive, non-competitive and uncompetitive (all reversible)
Reversible and irreversible inhibition

40
Q

Explain how competitive inhibition works

A

the inhibitor is chemically similar to the substrate, and there is competition for the active site between substrate and enzyme - can be overcome by increasing substrate conc.

41
Q

How is the slope and y-intercept affected in competitive inhibition with the addition of inhibitor? How does this relate to Km and Vmax?

A

slope - steeper
y-intercept - same

Km (apparent) is greater
Vmax remains the same

42
Q

Explain how non-competitive inhibition works

A
  • Inhibitors can bind to the enzyme at a site distant from the active site
  • Bind to ES or E
  • Non-competitive inhibition reduces kcat (i.e the maximum number of chemical conversions of substrate molecules per second)
43
Q

How is the slope and y-intercept affected in non-competitive (mixed) inhibition with the addition of inhibitor? How does this relate to Km and Vmax?

A

slope - steeper
y-intercept - higher

Km (apparent) is same (inhibitor doesn’t affect binding of enzyme to substrate, just lowers the concentration of usable enzyme)
Vmax is reduced

44
Q

Explain how uncompetitive inhibition works

A
  • Inhibitor binds only to the enzyme-substrate complex
  • Inhibitor binding site only accessible once substrate has bound
  • distort the active site to prevent the enzyme from being catalytically active
  • Cant have both bound at once
45
Q

How is the slope and y-intercept affected in uncompetitive inhibition with the addition of inhibitor? How does this relate to Km and Vmax?

A

slope - same
y-intercept - changes - lower

Km - reduced
Vmax - reduced

46
Q

Define Kcat

A

Vo = Kcat (ES)

kcat - rate law of ES to E+P

The rate of product formation under optimum conditions (i.e enzyme saturated with substrate) - maximum rate of velocity

47
Q

What is 1/Kcat

A

the time required to turn over or convert one molecule of substrate to product

48
Q

What is the most efficient way to determine enzyme efficiency?

A

Kcat/KM

Kcat tells us how many substrate molecules are converted into product by a single active site per unit time

KM describes the affinity of the substrate to the active site

49
Q

What is meant by steady state condition assumption?

A

ES complex concentration is not changing overtime. It remains constant, despite reactants and products concentrations changing over time

does not change because RATE IF FORMATION OF ES IS EQUALT OT THE THE RATE OF DISSOCIATION OF ES

50
Q

steady state assumption

How many reactions describe the rate of formation of ES?

A

E + S ⇋ ES (rate constant ka)

rate law: ka [E][S]

51
Q

steady state assumption

How many reactions describe the rate of dissociation of ES?

A

two equations

  1. ES ⇋ E + P
  2. ES ⇋ E + S

rate law:

kd [ES] + kcat [ES]

52
Q

What is the overall formula of the steady state assumption that leads to the formula for KM?

A

ka [E][S] = kd [ES] + kcat [ES]

(kd + kcat) / ka = [E][S]/[ES]

KM = [E][S]/[ES]

thus [ES] = [E][S]/KM

53
Q

How does [ES] = E][S]/KM relate to Vo? What does this tell us?

A 
Vo = Kcat [ES]
[ES] = [E][S]/KM

thus Vo = Kcat/ KM x [E][S]

The rate at which the enzyme catalyses. It tells us how enzymes operate under normal physiological conditions. That is when [S] «< KM

54
Q

Low Km is ___ affinity

A

high

55
Q

what is the best ratio for enzyme efficiency

A

low KM

high Kcat

56
Q

What is the limit of Kcat/KM

A

the highest value kcat/km can reach is equal to Ka.

57
Q

What does the michaelis menten equation allow us to calculate?

A

Vmax and Km

58
Q

What can a secondary plot tell us?

A

Determination of Ki

59
Q

What type of enzyme inhibitors work best as drugs?

A

tightly binding inhibitors of key enzymes
reversible inhibitors with a Ki value in the nM range
irreversible inhibition using mimetics or covalent binding

60
Q

What factors affect enzyme activity?

A

temperature - thermal denaturation

pH - dependence of a typical enzyme-catalysed reaction with ionisable amino acid(s) in the active site

61
Q

Describe how the reaction rate curve of an enzyme catalysed reaction differs between a single ionisable amino acid in the active site vs two

A
  1. s shape

2. hump

62
Q

What do you call an inorganic cofactor? Describe their function

A

activator - metal ions that may attach to active site to make its shape more efficient

63
Q

What is the main bond interaction of reversible inhibition?

A

hydrogen bonds

64
Q

Define irreversible inhibitors

A

attach to enzymes with strong covalent bonds which are difficult to break without damaging the enzyme. the effect of an irreversible inhibitor is permanent

65
Q

what is end product inhibition?

A

the end product of a reaction can act as a non competitive inhibitor controlling a series of enzyme catalysed reactions.

66
Q

What is an example of end-product inhibition?

A

negative feedback

67
Q

What is the name of the enzyme that is inhibited which functions as a regulatory enzyme in the metabolic pathway?

A

allosteric enzyme

68
Q

What is an allosteric enzyme

A

enzymes that have an additional binding site for effector molecules other than the active site.

69
Q

What is the homotropic effect in allosteric enzymes?

A

the binding of substrate to one active sire changes the affinity of the other active sites for substrate

70
Q

What is the difference between homotropic and heterotropic allosteric enzymes ?

A

Homotropic - communication between active sites such that the binding of substrate to one active site influences the further binding of substrate molecules to remaining active sites

Heterotropic - rate of reaction responds to the presence of substance unrelated to substrate. these effectors can be activators or inhibitors –> feedback loop

71
Q

What do the T and R states indicate in homotropic allostery?

A

T state - high Km - low affinity for S

R state - low Km - high affinity for S

Biosci 106 (2 decks)

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