Lecture 12 - How fast is that enzyme? continued Flashcards

1
Q

Michaelis-Menton equation

A
Vmax [S] 
V = ––––––––– 
       KM + [S]
=    [ES]
    \_\_\_\_\_\_\_\_
     [E] total
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2
Q

Lineweaver-Burk Plot

A

Fitting a curve to the Michaelis-Menten equation requires a computer.
Vmax [S]
V = –––––––––
KM + [S]

The double reciprocal plot can be fit to a line.
1 = KM ́ 1 +. 1
— — x — + —
V Vmax [S] Vmax

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3
Q

Interpretation of Km

A

Characterises one enzyme-substrate pair (if an enzyme can act on different substrates, it will have different KM values for each).
Is the substrate concentration needed to reach half Vmax - Units are units of concentration, e.g. mmol/ L
k -1 + k 2
Formally, KM = –––––––
k1

For many enzymes k 2 &laquo_space;k -1 , so approximation neglects k 2 :
k -1
KM = –––– (i.e. the ES dissociation constant)
K1
Low KM = high affinity between E and S; high KM = low affinity.

Km is the ratio of rate constants

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4
Q

Physiological significance of Km

A

In the cell, for a particular enzyme-substrate interaction, [S] is often below the KM.
• This means rate will rise to accommodate more substrate, tending to maintain steady state.

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5
Q

Km : substrate preference and response

A

Hexokinase generates energy in muscle and brain. Glucose + ATP → glucose-6-phosphate + ADP
KM for each isozyme and substrate is different.
Isozyme glucokinase stores energy as glycogen in the liver.

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6
Q

Kcat

A

Is the number of substrate molecules converted to product, per enzyme, per unit of time, when E is saturated with substrate. 

Therefore helps to define the activity of one enzyme molecule – a measure of catalytic activity. 

If the Michaelis-Menten model fits , kcat = k2 

Thus, kcat describes the ‘rate limiting’ step. 

Vmax = kcat [E]T
K2 = Vmax/ [E]T = Kcat

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7
Q

Kcat, Km and catalytic efficiency

A

The most effective enzymes should have…
• A high kcat (ability to turnover a lot of substrate into product, per second). Higher rate constant therefore can run fast
• A low KM (low substrate concentration required to achieve near Vmax; high affinity for the substrate under the Michaelis-Menten assumptions).
So kcat / KM is an overall measure of enzyme efficiency; the higher the better.

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8
Q

Kcat, Km and “catalytic perfection”

A

The upper limit for kcat / KM is the diffusion- controlled limit; i.e. the rate at which enzyme and substrate diffuse together. 

Viscosity of water sets an absolute upper limit at ~109 s-1 M-1. 

Enzymes with kcat / KM above 108 s-1 M-1 are referred to as ‘perfect’ catalysts.

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9
Q

Enzymes are optimised for …..

A

Specific roles

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10
Q

Enzyme inhibitors

A

Inhibitor: a compound that binds to an enzyme and reduces its activity. 

Important because:
o Natural inhibitors regulate metabolism.
o Many drugs, poisons & toxins are enzyme inhibitors. o Used to study enzyme mechanisms.
o Used to study metabolic pathways.

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11
Q

Two classes of inhibitors

A

Irreversible inhibitor – binds covalently to the enzyme.

Reversible inhibitor – not covalently bound to the enzyme.

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12
Q

Irreversible inhibitor

A

binds covalently to the enzyme.

Binds and ‘kills’ the enzyme and main effect you measure is that the amount of enzyme drops

Inhibitor binds to the enzyme and permanently inactivates it.
Inhibitor reacts with a specific amino acid side chain, usually in the active site, and forms a covalent bond

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13
Q

Irreversible inhibition

A

Inhibitors bind covalently to E, thereby inactivating them irreversibly (e.g. natural toxins)

Note: They can be competitive (most common) or non-competitive (less common) but in either way their binding will result in changing the enzyme’s structure so that it no longer works. Irreversible inhibitors bind permanently to their target enzyme, often via a covalent bond that influences catalysis. “Permanently” here means over a time-scale that is long compared to the functional lifetime of the enzyme itself (and that timescale may be minutes for some bacterial enzymes, and months or years for enzymes found in stationary populations of cells in eukaryotes). So, irreversible inhibition is brought about by covalent bonds, however, normal enzymatic reactions can also include convalent bonds. The covalent bonds in normal enzymatic reactions are however easily broken simply due to the enzymes being equipped to deal with that covalent bond. Phophorylation being one example and you’ve also been told about covalent catalysis (which includes a covalent bond being formed, but also broken).

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14
Q

Reversible inhibitor

A

not covalently bound to the enzyme.

A reversible inhibitor binds to the enzyme but can subsequently be released, leaving the enzyme in its original condition.

Inhibitor binds with non-covalent bonds therefore can let go of the enzyme

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15
Q

A reversible inhibitor may be either ….

A

Competitive or non-competitive

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16
Q

Covalent inhibitors often react with catalytic residues

A

Addition of the bulky tosyl-L-phenylalanine methylketone to the histidine disables the catalytic triad and fills the active site, blocking substrate binding.

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17
Q

Competitive inhibition

A

Inhibitor competes with the substrate for binding in the active site.

Michaelis-Menten Plot - No change in Vmax (looks like it is getting to the same asymptote but it just takes longer to get there (one of them takes longer), this is what happens with a competitive inhibitor because the inhibitor and substrate are wanting to bind in the same place): High [S] outcompetes the inhibitor. (if you have more substrate then there is going to be a majority more ES complexes

Lineweaver-Burk Plot - Increases KM (because of more substrate): More substrate is needed to get to V= Vmax /2.

  • Inhibitor (I) competes with the S for the active site on E
  • No reaction when I bound as S cannot access the active site - I binding does affect S binding - I is usually structurally AND chemically very similar to S and forms an EI complex
- Inhibition can be overcome by increasing [S]
  • Vmax is unchanged and Km increases which means that the enzyme can’t bind so well anymore
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18
Q

Transition state analogues as drugs

A

Enzymes are often targets for drugs. 
Transition state analogues can make ideal enzyme inhibitors. 
Enalapril and Aliskiren lower blood pressure. 
Statins lower serum cholesterol. 
Protease inhibitors are AIDS drugs. 
Juvenile hormone esterase is a pesticide target. 
Tamiflu is an inhibitor of influenza neuraminidase.

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19
Q

Transition state analogs make tight binding inhibitors

A

Adenosine deaminase uses a tetrahedral intermediate. 


A non-reactive analog, 1,6-dihydroinosine, effectively inhibits the enzyme.

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20
Q

Substrate analogue inhibitors of HIV protease

A

The inhibitor (shown as solid balls) fills the active site. 
Two catalytic aspartic acid residues are shown as pink rods below.

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21
Q

Non-competitive inhibition

A

Inhibitor binds at a different site than the substrate; it is an allosteric inhibitor. 
Enzyme can bind substrate, or inhibitor (I), or both. 
In pure non-competitive inhibition, the binding of I has no effect on the binding of S; i.e. the
substrate binds to E and EI with the same affinity.

Substrate binds just as well to the inhibiting complex and so the Km stays the same as you don’t need any more substrate to get to Vmax but you have a lower Vmax because there is less active enzyme since the inhibited stuff is not changing over

22
Q

Pure non-competitive inhibition

A

Binding I changes the structure of the active site such that S still binds, but transition state stabilisation is no longer optimal. —> Vmax decreases; KM stays the same.

I binds to other (i.e. allosteric) site on E than S (i.e. not in the active site)

  • I binding doesn’t effect S binding - E cannot work when I is bound
  • Vmax decreases and the Km remains unchanged (the ability to get into the active site will remain unchanged if the distortions to the other parts of the enzyme have no effect on the active site)
23
Q

Mixed non-competitive inhibition

A

More commonly, binding of the inhibitor does affect binding of the substrate —> mixed non-competitive inhibition. —> Vmax decreases; KM increases.

  • I binds to other site on E than S (i.e. not in the active site)
  • I binding does effect S binding - E cannot work when I is bound
  • Vmax decreases( due to structural deformities); KM increases.
24
Q

Competitive inhibitors may react

A

An alternate substrate can compete for the active site of an enzyme, as for alcohol dehydrogenase, on the right.
Typically, related molecules fail to react, but some do.

25
Q

Both competitive and non-competitive inhibition are …

A

Reversible because they don’t bind the inhibitor covalently (strong bonds) to the enzyme

26
Q

Allosteric enzymes in enzyme kinetics

A

Don’t follow Michaelis-Menten Kinetics

V vs [S] curve has an S-shape (sigmoidal) rather than being hyperbolic

27
Q

Catalytic rate constant

A

Kcat
How much S (in moles) is converted to P (in moles) per E (in moles) per unit time when enzyme is saturated with substrate

28
Q

Enzyme efficiency

A

How quickly the bound S can be converted to P (Kcat) relative to how well the E binds the S (Km)

Enzyme efficiency = Kcat/Km

kcat/KM results in the rate constant that measures catalytic efficiency. This measure of efficiency is helpful in determining whether the rate is limited by the creation of product or the amount of substrate in the environment.

A good enzyme is one that goes fast (has a high kcat) and binds well (low Km)

29
Q

Activators stabilise the _____ state

A

R state - (‘high affinity for the molecule it binds (substrate) configuration’) -In the presence of an activator, enzymes show little cooperatively as they are already in the high affinity configuration, so ‘team work’ is not really required to achieve high affinity for its substrate, hence the curve is more hyperbolic/ less sigmoidal and shifts to the left

30
Q

Inhibitors stabilse the _____ state

A

T state - (‘low affinity for molecule it binds (substrate) configuration’) - In the presence of an inhibitor, enzyme show more cooperatively as they are not in the high affinity configuration, so ‘team work’ is important to achieve high affinity for its substrate, hence the curve is less hyperbolic/more sigmoidal and shifts to the left

31
Q

R state

A

Is the configuration of any protein when it has high affinity for the substrate

32
Q

Why do enzymes increase the rate at which a specific substrate is converted to its products? - They provide substrate specificity

A

The 3D geometry of the active site provides substrate specificity

33
Q

Why do enzymes increase the rate at which a specific substrate is converted to its products? - They provide substrate binding affinity

A

The amino acid side chains that are part of the active site and interact with the substrate and participate in the chemical reaction provide the substrate binding affinity

34
Q

Why do enzymes increase the rate at which a specific substrate is converted to its products? - They facilitate and stabilise the transition state

A

Provides a much lower activation energy compared to the uncatalysed reaction

When the active site of the enzyme binds the substrate in a particular orientation and close to catalytic chains, the transition state that is required for the chemical reaction is facilitated and stabilised.

35
Q

Why do enzymes increase the rate at which a specific substrate is converted to its products? - They lower the (free) energy required to reach the transition state

A

As a result of the facilitation and the stabilisation of the transition state, the free energy required to reach the transition state is lowered

36
Q

When there is an excess of substrate, reaction velocity is _________ to enzyme concentration.

A

Proportional

37
Q

Why does an enzyme catalysed reaction reach a maximum velocity when there is excess substrate but a fixed concentration of enzyme?

A

By increasing the enzyme concentration, the maximum reaction rate greatly increases. Conclusions: The rate of a chemical reaction increases as the substrate concentration increases. Enzymes can greatly speed up the rate of a reaction. However, enzymes become saturated when the substrate concentration is high.

38
Q

What equation can be used to describe a V vs [S] curve for a monomeric enzyme?

A

Michaelis-Menten equation
Km is the Michaelis-Menten constant which shows the concentration of the substrate when the reaction velocity is equal to one half of the maximal velocity for the reaction. It can also be thought of as a measure of how well a substrate complexes with a given enzyme, otherwise known as its binding affinity.

39
Q

What are allosteric enzymes?

A

Allosteric enzymes are enzymes that change their conformational ensemble upon binding of an effector, which results in an apparent change in binding affinity at a different ligand binding site

40
Q

Ligand

A

a ligand is an ion or molecule (functional group) that binds to a central metal atom to form a coordination complex.

41
Q

Zymogen

A

an inactive substance that is converted into an enzyme when activated by another enzyme

42
Q

What does a high KM for an enzyme-substrate pair indicate?

A

An enzyme with a high Km has a low affinity for its substrate, and requires a greater concentration of substrate to achieve Vmax.”

43
Q

What does Kcat for an enzyme catalysed reaction represent?

A
  • Is the number of substrate molecules converted to product, per enzyme, per unit of time, when E is saturated with substrate.
  • Therefore helps to define the activity of one enzyme molecule - a measure of catalytic activity.
  • If the Michaelis-Menten model fits , kcat = k2
  • Thus, kcat describes the ‘rate limiting’ step.
44
Q

Can we accurately measure Vmax on a V vs [S] curve? Why/why not?

A

The double-reciprocal (also known as the Lineweaver-Burk) plot is created by plotting the inverse initial velocity (1/V0) as a function of the inverse of the substrate concentration (1/[S]). The Vmax can be accurately determined and thus KM can also be determined with accuracy because a straight line is formed.

45
Q

Linweaver-Burk plot

A

The double-reciprocal (also known as the Lineweaver-Burk) plot is created by plotting the inverse initial velocity (1/V0) as a function of the inverse of the substrate concentration (1/[S]).

46
Q

What does the y-intercept represent on a Lineweaver-Burk plot?

A

V

47
Q

What does the x-intercept represent on a Lineweaver-Burk plot?

A

S

48
Q

For what purpose are transition state analogues designed?

A

Transition state analogs can be used as inhibitors in enzyme-catalyzed reactions by blocking the active site of the enzyme. Theory suggests that enzyme inhibitors which resembled the transition state structure would bind more tightly to the enzyme than the actual substrate.

49
Q

How can competitive inhibition of an enzyme be overcome?

A

Competitive inhibition can be overcome by adding more substrate to the reaction, which increases the chances of the enzyme and substrate binding

50
Q

Where on the enzyme does the competitive inhibitor bind?

A

On the active site