Session 4.4a - Lecture 2 - Enzymes Flashcards

Slides 1 - 28

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1
Q

Enzyme activity: kinetics and inhibition

  • Role of enzymes
  • How enzymes work
  • Why they’re important
  • How we can study enzymes

Enzyme activity:

  • kinetics
  • measurement
  • inhibition
  • measuring inhibition
A

Role of enzymes (start to introduce what proteins do)

Answer these for ILO.

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2
Q

What are enzymes?

A

Biological catalysts that increase the rate of chemical reactions by lowering the activation energy.

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3
Q

What is a chemical reaction simplistically?

A

A chemical reaction is where we’re starting with a substrate and ending with a product. In many cases this can be REVERSIBLE, we can go in the opposite way.

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4
Q

Which normally has lower energy - the substrate or the product?

A

The product

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5
Q

What is the energy change for reactions commonly measured in?

A

Gibbs free energy (the change in free energy)

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6
Q

What causes the energy change in substrate to product creation?

A

Making and breaking bonds

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7
Q

When we look at energy diagrams there is a barrier we need to get across. What is this known as?

A

The activation energy

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8
Q

What is the activation energy?

A

The minimum amount of energy a substrate molecule must have to form our product.

“Minimum energy S must have to allow reaction”

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9
Q

What is the transition state?

A

The state of a chemical midway between the substrate and product.

“High energy intermediate that lies between S and P”

(e.g. think about breaking a stick in two - the original stick is the substrate; when it is flexed in half to break it is the transition state; the final two pieces are the product)

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10
Q

Draw a simple chemical reaction.

A

S P

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11
Q

Fig. 2

Label this graph.

A

x-axis: Progress of the reaction –>
y-axis: Free energy, G

beginning: S Ground state
end: P Ground state
middle: Transition state (+)

  • -> delta G+ S–>P (how much energy it takes to go from S to P)
  • -> delta G+ P–>S how much energy it takes to go from P to S)
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12
Q

Draw an energy diagram for an enzyme.

A

See Fig. 2

x-axis: Progress of the reaction –>
y-axis: Free energy, G

beginning: S Ground state
end: P Ground state
middle: Transition state (+)

  • -> delta G+ S–>P (how much energy it takes to go from S to P)
  • -> delta G+ P–>S how much energy it takes to go from P to S)
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13
Q

What happens to the rate of chemical reaction without a catalyst?

A

The rate of chemical reaction will not be very fast as not many molecules will have the activation energy.

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14
Q

How can you increase the rate of reaction?

A
  1. Temperature

2. Concentration

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15
Q

Why does increasing the temperature increase the rate of reaction?

A

It increases the number of molecules with activation energy

  • We give the molecules more energy so more of them are more likely to have the activation energy so they can overcome this intermediate state.
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16
Q

Why does increasing the concentration increase the rate of reaction?

A

Increases chance of molecular collisions

  • pack in more molecules, if there’s more of them there’s more chance of them interacting so more chance for reaction to occur
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17
Q

Why is it difficult to use temperature to increase the rate of reaction in humans?

A

If I took you and heated you up to 50 degrees, your reactions might occur faster, but then you’d also be dead …

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18
Q

Why is it difficult to use concentration to increase the rate of reaction in humans?

A

E.g. glycolysis

  • usually a few mM glucose in the cell
  • could whack it up to 1 M glucose
  • be much faster BUT
  • much more serious problems because cells would explode
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19
Q

Why do cells explode if you take in too much glucose?

A

Because the cells would take in so much water and hence explode

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20
Q

Why can’t we change temperature or concentration to increase the rate of reaction in a biological system?

A

The cells/you would die due to HOMEOSTASIS

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21
Q

What is the idea of homeostasis?

A

To keep things as constant as possible (yes, there are small changes in concentration, metabolism etc., but not by many orders of magnitude).

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22
Q

If we can’t increase the rate of reaction by temperature and concentration, how can we?

A

By binding catalysts, or enzymes.

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23
Q

How do enzymes increase the rate of reaction?

A

They lower the activation energy (catalysed reactions over uncatalysed reactions)

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24
Q

What does lowering the activation energy do in reactions?

A

It means more molecules are likely to have enough energy to react - this is the basis of what an enzyme it.

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25
Q

How does an enzyme lower the activation energy?

A
  • By facilitating the formation of a transition state
  • By bringing together molecules, e.g. if you get two reactants and you want them to react then by bringing them physically closer together, we increase
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26
Q

Fig. 3

Label this image

A

x-axis: Reaction progress –>
y-axis: Free energy –>

Substrate (beginning)
Transition state, S+ (hump/peak)
Product (end)

delta G+ (uncatalysed) (from S to higher S+)
delta G+ (catalysed) (from S to lower S+)
delta G for the reaction (substrate to product)

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27
Q

Draw the energy diagrams for a catalysed and uncatalysed reaction.

A

See Fig. 3

x-axis: Reaction progress –>
y-axis: Free energy –>

Substrate (beginning)
Transition state, S+ (hump/peak)
Product (end)

delta G+ (uncatalysed) (from S to higher S+)
delta G+ (catalysed) (from S to lower S+)
delta G for the reaction (substrate to product)

Reaction is the same but catalysed transition state must be LOWER than the uncatalysed transition state.

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28
Q

What are 6 important features of enzymes?

A
  • Highly specific
  • Unchanged after the reaction
  • Do not affect the reaction equilibrium
  • Increase the rate of reaction
  • Proteins
  • May require associated cofactors
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29
Q

Explain the phrase “highly specific” in relation to enzymes.

A

Most enzymes will only react with one certain substrate - they are very fussy and will not go after lots of different things.

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30
Q

What is the clinical significance of an enzyme being highly specific?

A

Normally, enzymes will only react with one substrate.

In some cases they will react with other substrates if things go wrong in the body, and you’ll do that in some units you do later, e.g. in galactosaemia for metabolism

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31
Q

Explain the phrase “unchanged after the reaction” in relation to enzymes.

A

Biological catalysts are unchanged after the reaction – might change during the reaction, might bind things, might move things around, often move protons in there - but at the end, enzyme will be same as the start

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32
Q

Identify four ways an enzyme can change DURING the reaction.

A

Enzyme might:

  • change during the reaction
  • bind things
  • move things around
  • often move protons in there

but ultimately, will remain unchanged at the end.

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33
Q

Explain the phrase “do not affect the reaction equilibrium” in relation to enzymes.

A

If the enzyme is unfavourable for certain enzymes it WON’T make it more likely to occur, it only increases the RATE.

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34
Q

Enzymes increase the rate of reaction.

What is a critical thing to remember about this?

A

They increase BOTH the forward and reverse reaction.

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35
Q

Are all enzymes proteins?

A

Most enzymes are proteins; not every but almost all, thus, virtually all enzymes are proteins.

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36
Q

What are cofactors?

A

Some enzymes may require associated help, from cofactors and coenzymes, to help them work, e.g. in metabolism.

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37
Q

Why are we interested in enzymes?

A
  • Inheritable genetic disorders
  • Overactive enzymes can cause disease
  • Measurement of enzyme activity for diagnosis
  • Inhibition of enzymes by drugs
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38
Q

What is the clinical significance of enzymes in inheritable genetic disorders?

A

Many mutations we see in inheritable genetic disorder often affect enzymes, e.g., many inheritable metabolic diseases have mutations involving enzymes.

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39
Q

What is the clinical significance of enzymes and disease?

A

By changing enzyme function (via mutation) you can cause disease.

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40
Q

How can overactive enzymes cause disease?

A

By switching on enzymes that are not normally present or at a certain level you can change the activity of an enzyme, which causes disease.

having an understanding of these enzymes and measuring them important for understanding disease properties.

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41
Q

Give a clinical example of overactive enzymes causing disease.

A

Cancer - signalling molecules have enzymes involved - things like tyrosine kinases are upregulated in many aspects of cancer.

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42
Q

What can the measurement of enzyme activity be used for clinically?

A
  • important for understanding disease properties

- for diagnosis, e.g. via reference ranges.

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43
Q

What can the measure of enzyme activity tell you about the enzyme itself?

A
  • Whether it’s being affected

- Whether an enzyme is in the right place - this will tell you whether the tissue is being damaged or not.

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44
Q

How do drugs target enzymes?

A

Many diseases are treated by drugs that target enzymes by acting as inhibitors.

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45
Q

What is the active site?

A

The active site of an enzyme is the place where substrates bind and where the chemical reaction occurs

(critical part of enzyme that does anything, point where substrates bind and reactions occur)

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46
Q

Describe the active site in relation to the size of the enzyme.

A

Active sites only occupy a relatively small part of the protein molecule (enzyme) itself

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47
Q

How many amino acids are active sites?

A

Most enzymes are >100 aa but active site is only a few aa

E.g. lysozyme, active site is only made up of ~6 amino acid residues (AAR) out of 129 AAR proteins (that’s quite a lot - some proteins are enormous but active site might be only a few amino acids)

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48
Q

Active sites only occupy a relatively small part of the enzyme itself.

Why do we need the rest of it then?

A

Most of enzyme acts as a scaffold to create the active site - enzyme works together to form a unique 3D shape to bring the active site together

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49
Q

Which amino acids in a sequence form the active site?

A

The active site is formed by amino acids from different parts of the primary sequence - so even though these residues are spread throughout the molecule, physically they’re close - forming an active site region where we can get binding of the substrate..

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50
Q

Fig. 5

Label and caption this diagram

A

Lysozyme
N 1 35 52 62,63 101 108 129 C

129 amino acid residues, 6 amino acid residues (numbered) form the active site.

(would not need to draw)

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51
Q

What shape are active sites?

A

Active sites are often clefts or crevices

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52
Q

Where is the active site of an enzyme?

A

Usually not on the surface of the protein, but more buried away

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53
Q

What is the significance of active sites being clefts or crevices?

A

Substrate molecules are bound in a cleft or crevice that usually
excludes water

Substrate molecules have to go into a cleft or crevice - important because defines what can get in, but in many cases used to exclude water

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54
Q

Substrate molecules are bound in a cleft or crevice that usually excludes water.

Why?

A

Water is so abundant and present at such high concentrations that it can interfere with many chemical reactions. It is important to exclude water in many cases, hence why you get active sites in these unique shapes (cleft/crevices).

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55
Q

What is the relationship between the shape of the active site and the substrate?

A

Active sites have a complementary shape to the substrate

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56
Q

What enzyme theory came out at the first half of the 20th century explaining active site complementarity?

A

“Lock and Key” hypothesis

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57
Q

What is the “Lock and Key” hypothesis and what is it’s potential problem?

A

Idea that the active site is complementary in shape to the substrate – substrate can come in and bind – then get a enzyme-substrate complex which then can be released.

But if a substrate is coming in and binding to the enzyme, why would it want to let go again - so it doesn’t sort of fully make sense.

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58
Q

Fig. 6

Label the image

A

Substrate + Enzyme Active site a b c –> ES complex a b c

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59
Q

Draw a reaction between an enzyme and a substrate from the “Lock and Key” hypothesis.

A

See Fig. 6

Substrate + Enzyme Active site a b c –> ES complex a b c

Substrate shape must be complementary to the enzyme, and fit into the enzyme perfectly to form an ES complex.

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60
Q

What is the “Lock and Key” hypothesis?

A

The active site of the enzyme is complementary in shape to that of the substrate.

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61
Q

After the “Lock and Key” hypothesis, what was the hypothesis used to explain enzyme-substrate binding?

A

“Induced fit” hypothesis

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62
Q

Does the “induced fit” hypothesis follow the rule that active sites have a complementary shape to the substrate?

A

The active site is complementary, but it only occurs after the substrate binds.

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63
Q

How can the “induced fit” hypothesis work alongside the “lock and key” hypothesis?

A

In the “induced fit” hypothesis, the active site is complementary, but it only occurs after the substrate binds.

This means there is some sort of complementary – e.g. particular amino acid residue that will bind substrate and only become important once physically bound – giving you “lock and key” hypothesis

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64
Q

Explain what we currently believe to be true about the active site of enzymes and their substrates.

A

Active sites have a complementary shape to the substrate
“Lock and Key” hypothesis

Binding of substrates can induce changes in the conformation
“Induced fit” hypothesis

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65
Q

When is the active site complementary to the substrate following the “induced fit” hypothesis?

A

The active site only forms a complementary shape AFTER binding of the substrate

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66
Q

Fig. 7

Label the image

A

Substrate + Enzyme a b c –> ES complex a b c

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67
Q

Draw enzyme-substrate binding in the “induced fit” hypothesis.

A

Substrate + Enzyme a b c –> ES complex a b c

Enzyme has a non-complementary but similar shape to the substrate. It becomes complementary after binding of the substrate.

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68
Q

How are substrates bound to the active site?

A

Substrates are bound to enzymes by multiple weak bonds - non-covalent interactions.

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69
Q

Why do substrates bind to enzymes with NON-covalent interactions?

A

We generally don’t want them to form strong covalent bonds – covalent bonds then they won’t get away again, so binding must not be too tight!

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70
Q

Substrates tend to bind to active sites through multiple weak bonds.

How can this be exploited pharmacologically?

A

Substrates don’t form strong bonds with active sites (such as covalent bonds) - otherwise they wouldn’t be able to detach.

However, inhibitors can be made which attach covalently to active sites. This means they effectively block the active site because the strong covalent bonds mean that they won’t leave the active site; so the substrate can’t attach (thereby inhibiting the active site).

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71
Q

How does uracil interact with ribonuclease?

A

Uracil, from a substrate, can bind to the enzyme ribonuclease. The side chain uracil makes several interactions with the active site residues from the ribonuclease enzyme.

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72
Q

What stabilises a uracil side chain when it interacts with a ribonuclease active site?

A

Hydrogen bond interactions from serine and threonine that stabilise uracil binding within active site, holding it in there whilst reaction is occurring

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73
Q

Fig. 8

Label the image

A

Uracil (from substrate)

Serine side chain

Threonine side chain

e.g. ribonuclease

(would not need to draw)

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74
Q

How do enzymes help form the transition state?

A

Enzymes hold the substrate in place to form the enzyme-substrate complex - this puts the substrate in the right place, forming the enzyme-substrate transition state, allowing the product to be formed.

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75
Q

How do enzymes help the substrate form a product? Describe using a parent-baby anology.

A

Baby (substrate) is wearing a jumper but wants to take it off because it’s too hot. By itself, the baby is very uncoordinated and flails around. It could take the jumper off if its arms are in the right direction but it needs a bit of help.

Our parent (presumably) is the enzyme which is able to hold the baby (S) in the right conformation, forming our ES complex.

The parent can guide the baby’s arms into the correct position to get the jumper off, to form the ES transition state. This is the nice state where the jumper can be taken off.

The jumper is taken off the baby (baby minus jumper) to form our product. Happy baby.

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76
Q

Fig. 9

Label this image

A

Substrate (S) –> Enzyme Enzyme-substrate complex (ES) –> Enzyme Transition state –> Product (P)

Graphs:
x-axis: Progress of reaction -->
y-axis: Free energy
curve
- S (beginning)
- ES (dip)
- T* (peak)
- P (end)
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77
Q

Draw an energy diagram explaining how enzymes work to lower activation energy.

A

See Fig. 9

x-axis: Progress of reaction -->
y-axis: Free energy
curve
- S (beginning)
- ES (dip)
- T* (peak)
- P (end)
  • S is the substrate
  • Enzymes holds substrate to form ES complex, which lowers activation energy
  • This allows T* (transition state) to form, which is the peak activation energy
  • Leaving with a product, P, at the end, with the lowest activation energy
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78
Q

What do we mean by enzyme kinetics?

A

This is really just measuring the activity of enzymes.

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79
Q

What happens to an enzyme as a chemical reaction occurs?

A

If we think simplistically about an enzyme and what’s going to happen as a chemical reaction occurs, need to think about in terms of concentration of substrate, product and amount of time that’s passed by.

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80
Q

Why is T=0 important in enzyme reactions?

A

If we prepare a reaction (we’ve put some substrate in a tube and put it in the right pH, right buffer) and add some enzyme at time 0, then this is the only point we actually know how much substrate there is - we know the substrate concentration because we put a defined concentration of substrate in.

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81
Q

Why is T=0 the only point we actually know how much substrate there is in a reaction?

A

We know the substrate concentration here because we put a defined concentration of substrate in. However, this changes as a reaction progresses

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82
Q

What happens to the concentration of substrates as the reaction progresses?

A

Enzyme will work and chemical reaction will occur and more of our substrate will turn into product, thus, over time, more product will appear.

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83
Q

How can enzyme reactions be measured?

A

By measuring the appearance of product over time
(substrate turns into product with time in a chemical reaction).

We can also measure substrate disappearing as well.

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84
Q

How do product-time reaction graphs tend to look?

A

Tend to see sort of curve showing reaction slowing over time – amount of substrate decreases over time and product increases.

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85
Q

What is V0?

A

The initial velocity or initial rate that is drawn as a tangent to the product-time curve.

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86
Q

Why is V0 important?

A

Only time we actually know concentration of substrate is right at the start – so V0 is drawn as a tangent to this curve to work out the rate of reaction: change in product over time.

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87
Q

Fig. 10

Label this image

A

T=0 - all substrate, no product (and enzyme)

T=10 - some substrate, some product (and enzyme)

T=20 - less substrate, more product (and enzyme)

Graph:
x-axis: Time –>
y-axis: Product –>
Tangent = V0

V0 = initial rate of reaction

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88
Q

Draw a diagram demonstrating enzyme kinetics

A

Fig. 10 (top)

T=0 - S S S S S S S S E
-->
T=10 - S S S S S P P P E
-->
T=20 - S S P P P P P P E
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89
Q

Draw a graph demonstrating enzyme kinetics

A

Fig. 10 (bottom)

x-axis: Time –>
y-axis: Product –>
Tangent = V0
V0 = initial rate of reaction

curve shape - hyperbola

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90
Q

How do enzyme kinetics vary with different properties?

A

Enzymes show different kinetics with different properties, e.g. temperature, pH.

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91
Q

What sort of curve does an enzyme make when looking at kinetics of different properties?

A

A bell-shaped curve, with a maximum in the middle and reaction rates falling off sharply on either side.

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92
Q

How does temperature affect enzymes in your body?

A

The maximal rate of reaction will be at 37oC, or body temperature.

The rate of reaction will dip if the temperature is lowered/increased either side of this - this in your body is not going to make too much difference because the enzymes are not really going to see much difference from around 37oC.

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93
Q

How does temperature affect non-human enzymes?

A

Other enzymes in different organisms have a different temperature maximal rate:

  • an enzyme taken from hot springs bacteria (important in PCR for thermostable polymerases) can be optimally active around 95 oC.
  • enzyme taken from a marine bacterium – sea temp normally around 4oC, cyrophilic (extremophilic organisms that are capable of growth and reproduction in cold temperatures) bacteria, temp optimum around about 4-5 oC. (Normally a range)
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94
Q

How is temperature used for enzymes involved in PCR?

A

An enzyme taken from hot springs bacteria (important in PCR for thermostable polymerases) can be optimally active around 95 oC.

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95
Q

What is normal sea temperature?

A

Sea temp normally around 4oC

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96
Q

What is a cryophilic organism?

A

Extremophilic organisms that are capable of growth and reproduction in cold temperatures

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97
Q

How does pH affect enzyme kinetics?

A

Like temperature, you’d see a bell-shaped curve.

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98
Q

What is the optimum pH for most enzymes in the human body?

A

Most human enzymes have a pH optimum about 7.4, bc this is the pH you’d see in most of your tissues or cells or organelles (not completely true bc certain organelles more acidic, so there might be enzymes that optimise for that)

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99
Q

What is the optimum pH for pepsin, and why?

A

In certain tissues as well such as the stomach, the pH ~1-2.

Pepsin is a protease present in the stomach which has a pH optimum around about 2, so it works very well in acidic environment (has a slightly different make up of course that will allow it to do that

100
Q

Fig. 11

Label the diagram

A
Temperature
x-axis: Temperature oC
4 37 95
y-axis: rate of reaction
red: human enzyme
pH
x-axis: pH 1 2 5 7
y-axis: rate of reaction
green: e.g. pepsin
red: e.g. carbonic anhydrase
101
Q

Draw a graph to show the effect on enzyme kinetics temperature has, in a human enzyme, a cryophilic enzyme and an enzyme involved in PCR.

A

Fig. 11 (top)

Temperature
x-axis: Temperature oC
4 37 95
y-axis: rate of reaction
red: human enzyme

bell shaped curved peaking at numbers shown, dipping either side

102
Q

Draw a graph to show the effect on enzyme kinetics pH has in an acidic and neutral enzyme. Give two examples.

A
pH
x-axis: pH 1 2 5 7
y-axis: rate of reaction
green: e.g. pepsin
red: e.g. carbonic anhydrase

bell-shaped curves, pepsin peaking at 2, carbonic anhydrase peaking around 7, dipping either side.

103
Q

How can we measure the affect a substrate concentration has on reaction rate?

A

We’ve seen that we can measure the appearance of product with time and get the reaction rate, V0. So we can do that with diff conc of substrates and plot a graph of reaction rate as a function of substrate concentration.

104
Q

What shape does a curve with a reaction rate as a function of substrate concentration have?

A

Rectangular hyperbola

105
Q

What is maximal velocity?

A

If we went to infinite substrate concentration we’d eventually reach a maximal velocity where the enzyme can’t work any faster (no matter if more substrate is added).

106
Q

Fig. 12

Label and caption the curve.

A

x-axis: Substrate concentration –>
y-axis: Reaction velocity –>
Dashed line: Maximal velocity

Reaction rate as a function of substrate concentration.
An enzyme-catalysed reaction reaches a maximum velocity: the plot often has the shape of a rectangular hyperbola

107
Q

Draw a graph of reaction rate as a function of substrate concentration (3 marks).

A

See Fig. 12

x-axis: Substrate concentration –>
y-axis: Reaction velocity –>
Dashed line: Maximal velocity

Axis (1 mark)
Maximal velocity (1 mark)
Rectangular hyperbola (1 mark)
108
Q

How do you represent the reaction rate as a function of substrate concentration mathematically?

A

Via the Michaelis-Menten equation

109
Q

What is the first equation used in the derivation of the Michaelis-Menten equation?

A

E + S –k1–>* ES –k2–> E + P

*

110
Q

Explain this equation

E + S –k1–>* ES –k2–> E + P
*

A

An enzyme can bind a substrate to form an ES complex that can either go on to form the product and the enzyme or it can fall apart again to release substrate and enzyme. There are different rate constants involved in all of this (k).

111
Q

What is an important consequence of this equation?

E + S –k1–>* ES –k2–> E + P
*

A
  • enzyme and substrate come together to form product or fall apart.
  • this gives you the Michaelis-Menten equation
112
Q

What is the Michaelis-Menten equation?

A

V0 = Vmax [S] / Km + [S]

Km = Michaelis constant (M)
Vmax = maximal velocity (mol/min)
113
Q

What are the constants in the Michaelis-Menten equation?

A

Km = Michaelis constant (M)

Vmax = maximal velocity (mol/min)

114
Q

What does the Michaelis-Menten equation tell us?

A

It predicts that a plot

of V0 versus [S] will be a rectangular hyperbola.

115
Q

What are the units of Km?

A

M (molar) or mol/L

units of concentration

116
Q

What are the units of Vmax?

A

mol/min (or mmol/min or umol/min)

units of rate

117
Q

What is a key thing to note about Km and Vmax?

A

THEY HAVE DIFFERENT UNITS!

Km = mol/L (M) (concentration)

Vmax = mol/min (rate)

118
Q

What is the Michaelis-Menten model for enzyme catalysis?

A

The model proposes that a specific complex between the enzyme and the substrate is a necessary intermediate in catalysis.

119
Q

What does this form the basis of?

E + S –k1–>* ES –k2–> E + P
*

A

The Michaelis-Menten model for enzyme catalysis

The model proposes that a specific complex between the enzyme and the substrate is a necessary intermediate in catalysis.

120
Q

What is Km?*

A

Substrate concentration that gives half maximal velocity

Do NOT say half the maximal velocity, as that implies it has unit of rate NOT concentration - be specific

(*has been previous exam question)

121
Q

What is Vmax?

A

Maximal rate when all enzyme active sites are saturated with substrate (definition)

122
Q

Explain what Vmax means.

A

It’s the maximal rate, as fast as an enzyme can possibly work -

Have an enzyme molecule, a substrate will come in, chemical reaction, substrate gets released. An enzyme that’s working as fast as it can is one where as soon as it’s done a chem reaction, it will be replaced by another substrate - it’s limited almost by diffusion; how fast you can get things in and out of the active site

123
Q

When does Vmax occur?

A

It is a theoretical maximal, rate - can only get it at infinite [S] (can get very close to it but not fully, but gives idea of how fast Vmax occurs)

124
Q

How do we work out Km?

A

The Michaelis constant Km is the substrate conc that gives half maximal velocity, so what we’re doing on graph

  • look at y-axis, here’s the max rate
  • take half the max rate
  • look across on the curve and follow down onto x-axis
  • that tells us our Km value

(Bc we’re reading on x-axis, Km is in units of substrate conc, and that’s why that has those units – so plz be specific – Km is NOT half Vmax, if you said that it would mean it’s got units of rate. Plz don’t stay kilometres)

125
Q

Fig. 14

Label the graph

A

x-axis: Substrate concentration [S] –>
y-axis Reaction velocity (V0) –>

Vmax (top line)
halfway down - Vmax/2
read off graph - Km

126
Q

Draw a graph showing the reaction velocity as a function of substrate concentration. Label Vmax and Km.

A

See Fig. 14

x-axis: Substrate concentration [S] –>
y-axis Reaction velocity (V0) –>

Vmax (top line)
halfway down - Vmax/2
read off graph - Km

127
Q

What is the maximal rate when all enzyme active sites are saturated with substrate?

A

Vmax

128
Q

What is the substrate concentration that gives half maximal velocity?

A

Km

129
Q

What is the significance of Km values?

A

Km values give a measure of the affinity of an enzyme for its substrate

130
Q

What does it mean if you have a low Km value?

A

Low Km = high affinity for substrate

131
Q

What does it mean if you have a high Km value?

A

High Km = low affinity for substrate

132
Q

Explain how having a low Km means you have a high affinity for substrate?

A

Low Km tells you the enzyme reaches half maximal rate at a v low conc - meaning it has high affinity.

133
Q

Explain how having a high Km means you have a high affinity for substrate?

A

If enzyme has high Km value it means it takes a lot longer to reach half maximal rate, hence a low affinity for substrate.

134
Q

What do hexokinase and glucokinase catalyse respectively?

A

They catalyse the same reaction

135
Q

How are hexokinase and glucokinase related?

A

They are isoenzymes: enzymes that catalyse the same reaction but have a slightly diff AA sequence

136
Q

What is hexokinase?

A

An enzyme that - catalyses the conversion of glucose to glucose-6-phosphate

  • in skeletal muscle
  • has a low Km (0.1 mM)
137
Q

What catalyses the conversion of glucose to glucose-6-phosphate in skeletal muscle?

A

Hexokinase

138
Q

What is glucokinase?

A
  • A type of hexokinase
  • catalyses glucose to glucose-6-phosphate
  • is only found in the liver
  • has a higher Km (relative to hexokinase) (5 mM compared to 0.1 mM)
139
Q

Hexokinase has a Km of 0.1 mM. What is the clinical significance of this?

A

At 0.1 mM glucose HK will have half maximal rate, so by the time you get to 1 mM it’s almost fully active. Important bc hexokinase is active all the time, bc if you’re taking glucose into cells, there’s going to be enough glucose to make sure it’s at max velocity all the time.

140
Q

Glucokinase has a Km of 6 mM. What is the clinical significance of this?

A

Glucokinase has Km of 5 mM – only becomes significantly active when you have >5 mM glucose. Normal conc of glucose in blood is around about 4-5 mM, if you think about gradients, in the liver, it’s not going to be that active most of the time, but when you eat, blood glucose levels go up - take more glucose into liver, glucokinase becomes more active, and that actually fits with metabolic function of liver – liver is going to use excess glucose to store as glycogen.

141
Q

What is the normal concentration of glucose in the blood?

A

About 4-5 mM

142
Q

How do HK’s and GK’s Km fit with their function?

A

HK is always active whereas GK only becomes active when glucose levels peak after feeding

143
Q

Fig. 15

Label this graph for hexokinase and glucokinase.

A

x-axis: [glucose] mM 0 5 10 15
y-axis: Rate 0.0 0.2 0.4 0.6 0.8 1.0
Red line: hexokinase
Blue line: glucokinase

144
Q

Draw a graph for hexokinase and glucokinase as a function of rate against [glucose] mM.

A

See Fig. 15

x-axis: [glucose] mM 0 5 10 15
y-axis: Rate 0.0 0.2 0.4 0.6 0.8 1.0
Red line: hexokinase
Blue line: glucokinase

Rate 1.0/0.5
HK - 1 mM/0.1 mM
GK - >15 mM/5 mM

145
Q

Fig. 15

Explain this graph.

A

Hexokinase is an enzyme that catalyses the reaction glucose to glucose-6-phosphate in skeletal muscle. It has a low Km of 0.1 mM which means it has a very high affinity for glucose. It reaches very close to Vmax at 1 mM, and as there is normally more than 1 mM glucose in cells, hexokinase is pretty much constitutively active.

Conversely, glucokinase is an isoenzyme of hexokinase with a Km of 5 mM. An isoenzyme it an enzyme that catalyses the same reaction but has a different amino acid sequence. Glucokinase is found only in the liver, as opposed to HK and skeletal muscle. It has a much higher Km compared to HK; a Km of 5 mM. Normal blood glucose is around 4-5 mM (non-diabetics) which means that glucokinase isn’t normally active. However, when you eat, your blood glucose increases, and more glucose is uptaken into the liver, meaning GK has a Km designed to fit its function: the increase of glucose >5 mM will activate GK so that glucose can be converted into glycogen in the liver for storage.

(This explains everything on Slide 15)

146
Q

2 enzymes have the following Km values:
Enz 1 = 50uM
Enz 2 = 75uM

Which has the highest affinity for its substrate?

A

Enz 1

Highest affinity = lowest Km

147
Q

2 enzymes have the following Km values:
Enz 1 = 53uM
Enz 2 = 0.053mM

Which has the highest affinity for its substrate?

A

Neither! They’re the same

(53 uM = 0.053 mM
Remember to convert your values then apply the same rules;
highest affinity = lowest Km)

148
Q

What does measuring Vmax/V0 in units do?

A

It standardises it

149
Q

What is 1 unit?

A

The amount of enzyme that converts 1umol of product per

min under standard conditions

150
Q

What are standard conditions?

A

25 oC and 1 M (but remember that in reality this is not normally the case, e.g. the temperature is normally 37 oC in the body)

151
Q

How are Vmax/V0 expressed?

A

Measured in amounts per unit time, and often expressed as a standardised rate.

152
Q

Why do we express Vmax/V0 at a standardised rate?

A

No point doing it at random amount, need to standardise it - there may be experimental factors that affect the rate, you might have:

  • extracted enzymes from tissue
  • different amt of tissue to start with
  • diff amounts of serum
  • different forms of extraction which are not as efficient as others
153
Q

What is the standardised rate?

A

per litre (L) of serum or per gram (g) of tissue

154
Q

What is the relationship between rate of reaction and enzyme concentration?

A

The rate of an enzyme catalysed reaction is proportional to the
concentration of enzyme

155
Q

Is the rate of an enzyme catalysed reaction proportional or inproportional to the
concentration of enzyme?

A

The rate of an enzyme catalysed reaction is proportional to the
concentration of enzyme

156
Q

Is the rate of an enzyme catalysed reaction proportional or inproportional to the
concentration of enzyme?

A

The rate of an enzyme catalysed reaction is proportional to the
concentration of enzyme

157
Q

The rate of an enzyme catalysed reaction is proportional to the
concentration of enzyme

Explain.

A

If you double the amount of enzyme - double the rate

i.e. if you’ve got one enzyme molecule and lots of substrate – you know it can work at a certain rate – it will produce so many product molecules per min. If you have 2 enzymes, they’ll work at twice the rate and produce twice as much

158
Q

The rate of an enzyme catalysed reaction is proportional to the
concentration of enzyme.

Explain the effect this has on the standardised rate.

A

Double the amount of enzyme - double the rate

BUT!!! - not the standardised rate

(bc if you have twice as much enzymes then you also have twice as much protein, so STANDARDISED rate is the same)

159
Q

The activity of enzyme in 1ml of serum was measured to be 7.7
nmol per min.

What is the activity of the enzyme in units per L?

A

7.7nmol/ml is equivalent to 7700nmol/L

7700nmol/L is equivalent to 7.7umol/L

Therefore, 7.7 units/L

(Remember 1 unit = 1 umol/min
Step 1 - remember to scale up by 1000
Step 3 - remember to specify in units)

160
Q

What is the Lineweaver-Burk plot?

A

The Michaelis-Menten equation can be rearranged to give a linear plot

161
Q

Why do we use the Lineweaver-Burk plot?

A

It is a simple way of representing enzymatic activity in a linear form rather than a curve.

162
Q

What is the Lineweaver-Burk plot equation?

For interest

A

Michaelis-Menten equation:

V0 = Vmax [S] / Km + [S]

So Lineweaver-Burk plot:
1/V0 = Km/Vmax . 1/[S] + 1/Vmax

(Remember MMe and that LK is rearrangement for 1/V0)

163
Q

What is on the x-axis of a Lineweaver-Burk plot?

A

1 / [S]

164
Q

What is on the y-axis of a Lineweaver-Burk plot?

A

1 / V

165
Q

What does the gradient of the Lineweaver-Burk plot tell you?

A

Km / Vmax

166
Q

What does the x-intercept (y=0) on the Lineweaver-Burk plot tell you?

A

-1 / Km

167
Q

What does the y-intercept (x=0) on the Lineweaver-Burk plot tell you?

A

1 / Vmax

168
Q

What is the function of the Lineweaver-Burk plot?

A

Allows for easy estimation of Km and Vmax from linear plot

169
Q

Fig. 19

Label this graph.

A
x-axis: 1 / [S]
y-axis: 1 / V
0 co-ordinates
Slope = Km/Vmax
Intercept (x-axis) = -1 / Km
Intercept (y-axis) = 1 / Vmax
170
Q

Draw a Lineweaver-Burk plot and explain what it shows you.

A

See Fig. 19

x-axis: 1 / [S]
y-axis: 1 / V
0 co-ordinates
Slope = Km/Vmax
Intercept (x-axis) = -1 / Km
Intercept (y-axis) = 1 / Vmax
171
Q

Fig. 20

Which enzyme shows the highest affinity for its substrate?

A

A - lowest Km value so highest affinity

(x-axis = -1/Km
so Km = -1/x-axis

A = -1/-9 = 0.1111111
B = -1/-3 = 0.33333

So A has the lowest number hence lowest Km, meaning highest affinity)

172
Q

Fig. 20

Which enzyme shows the highest Vmax?

A

B

(y-axis = 1/Vmax
Vmax = 1/y-axis

A: 1/50 = 0.02
B: 1/25 = 0.04

0.04 is higher so B has a higher Vmax)

173
Q

How can you compare Km’s on >1 Lineweaver-Burk plot?

A

The x-axis that is most to the left means it has the smallest Km.

174
Q

How can you compare affinity on >1 Lineweaver-Burk plot?

A

Affinity = measure of Km (the smaller the Km the higher the affinity)

The x-axis that is most to the left means it has the smallest Km, hence highest affinity.

175
Q

How can you compare Vmax on >1 Lineweaver-Burk plot?

A

The y-axis that is the lowest down has the largest Vmax.

remember it is reciprocal - the lower down it goes the higher Vmax

176
Q

How can you compare rate on >1 Lineweaver-Burk plot?

A

The higher the Vmax the quicker the rate.

The y-axis that is the lowest down has the largest Vmax.

177
Q

Fig. 20

Label these Lineweaver-Burk plots

A

x-axis: 1/[S] (mM)
-10 0 10

y-axis: 1/v (U/L)
0 100

178
Q

Draw 2 Lineweaver-Burk plots, with the left graph having higher affinity but the right graph having a higher Vmax.

A

See Fig. 20

x-axis: 1/[S] (mM)
-10 0 10
y-axis: 1/v (U/L)
0 100

x-intercept more negative on the LEFT graph but y-intercept more positive

y-intercept more negative on the RIGHT graph but y-intercept more positive

x-intercept must always be <0

179
Q

What are enzyme inhibitors?

A

Molecules that slow down or prevent an enzyme reaction

180
Q

Why do we need to know about enzyme inhibitors?

A

There are many different classes of inhibitors that you will come across - and many inhibitors act as drugs

181
Q

How do drugs inhibitors work?

A

They can lower a rate of enzyme activity, creating an effect on overall activity of enzyme – e.g. if activity is abnormal then it brings it back to normal system

182
Q

What types of inhibitors can we get?

A

1) Irreversible

2) Reversible
i) Competitive
ii) Non-competitive

183
Q

What is the mechanism of action of irreversible inhibitors?

A

Bind very tightly, generally form covalent bond(s).

184
Q

Give an example of an irreversible inhibitor?

A

Nerve gases, such as sarin

185
Q

What is sarin?

A

Sarin is a potent inhibitor of acetylcholinesterase, an enzyme that degrades the neurotransmitter acetylcholine after it is released into the synaptic cleft. It is used as a deadly nerve agent (chemical weapon).

186
Q

What is the mechanism of action of reversible inhibitors?

A

Non-covalent; can freely dissociate.

187
Q

What is the mechanism of action of competitive reversible inhibitors?

A

Binds at active site

188
Q

How do competitive reversible inhibitors affect enzyme kinetics?

A

Km increases (or apparent Km); Vmax unaffected

189
Q

What is the mechanism of action of non-competitive reversible inhibitors?

A

Binds at another site on the enzyme (other than the active site)

190
Q

How do non-competitive reversible inhibitors affect enzyme kinetics?

A

Km unaffected; Vmax decreases

191
Q

Describe irreversible and reversible enzyme inhibitors.

A

Irreversible: binds tightly via covalent bonds

Reversible: does not bind via covalent bonds; can freely dissociate

  • Reversible: binds at active site, affects Km but not Vmax
  • Irreversible: binds away from active site, affects Vmax but not Km
192
Q

How do competitive inhibitors work?

A

Substrates bind to enzyme active sites by their complementary interactions.

The competitive inhibitor resembles the substrate and binds to the active site, reducing the proportion of enzyme molecules bound to the substrate.

193
Q

Give an example of a competitive reversible inhibitor.

A

Tamiflu

194
Q

What is Tamiflu?

A

An antiviral medication for influenza (flu).

195
Q

What is the mechanism of action of Tamiflu?

A

It is an competitive reversible inhibitor of NEURAMINIDASE found in influenza (flu) viruses by resembling N-acetyl neuraminic acid.

196
Q

What does Tamiflu resemble?

A

N-acetyl neuraminic acid, for binding in influenza neuraminidase. (Has similar ring structure and ‘legs’)

197
Q

Tamiflu is the trade name. What is the chemical name?

A

Oseltamivir

198
Q

Oseltamivir is the chemical name for which drug?

A

Tamiflu (trade name).

199
Q

Fig. 22 (top)

Label this image depicting competitive inhibition.

A

Substrate - Enzyme

Competitive inhibitor - Enzyme

200
Q

Fig. 22 (bottom)

Label this image.

A

N-acetyl neuraminic acid

Oseltamivir (TAMIFLU)

201
Q

Draw an image depicting the mechanism for competitive inhibition.

A

See Fig. 22 (top)

Substrate - Enzyme
Competitive inhibitor - Enzyme

202
Q

Explain the kinetic properties in competitive reversible inhibition.

(Explains Slide 23)

A

Km is affected, Vmax unaffected.

Imagine you’ve got an enzyme, and you have 1 substrate molecule and 1 inhibitor molecule which bind at equal affinity, then 50:50 chance which one will go in – either could bind. If you have lots more inhibitor – more likely for inhibitor to bind to active site. If you add more substrate, you can outcompete inhibitor – so as you add more substrate, effect of inhibiton goes away. Infinite substrate concentration means you completely abolish it.

So with an inhibitor, the more substrate you add then the effect of inhibition is reduced because it outcompetes, hence no reduction in Vmax. But there will be a reduction in Km, because more substrate will be needed to bind to the enzyme, as there is competition from the inhibitor.

203
Q

Why does competitve inhibition not have an effect on Vmax?

A

Adding enough substrate will always overcome the effect of the inhibitor, so no effect on Vmax

204
Q

Why does competitve inhibition not have an effect on Km?

A

Because the inhibitor competes with the substrate for the active site Km increases

205
Q

Km increases; Vmax unaffected.

What is this describing?

A

Competitive inhibition

206
Q

Fig. 23 (left)

Label this graph.

A

x-axis: [Substrate] –>
y-axis: Relative rate 0 20 40 60 80 100

Lines from top to bottom
Black: No inhibitor
[I] = Ki
[I] = 5 Ki
[I] = 10 Ki

(i.e. the Km increases with the more inhibitor
Ki = inhibition and it increases by rate with increased substrate
Vmax will eventually stay the same)

207
Q

Fig. 23 (right)

Label this graph.

A

x-axis: 1/[S]
y-axis: 1/V
0 coordinates

black: No inhibitor present
pink: + Competitive inhibitor

(Vmax is the same [y-axis] but Km is increased with competitive inhibition [x-axis more positive])

208
Q

Draw a Michaelis-Menten plot of enzyme activity with 3 competitive inhibitors.

A

See Fig. 23 (left)

x-axis: [Substrate] –>
y-axis: Relative rate 0 20 40 60 80 100

Lines from top to bottom
Black: No inhibitor
[I] = Ki
[I] = 5 Ki
[I] = 10 Ki

(i.e. the Km increases with the more inhibitor
Ki = inhibition and it increases by rate with increased substrate
Vmax will eventually stay the same)

209
Q

Draw a Lineweaver-Burk plot of an enzyme with and without a competitive inhibitor (3 marks)

A

See Fig. 23 (right)

x-axis: 1/[S]
y-axis: 1/V
0 coordinates (1 mark)

black: No inhibitor present
pink: + Competitive inhibitor

(Vmax is the same [y-axis] (1 mark)
but Km is increased with competitive inhibition [x-axis more positive] (1 mark))

210
Q

Explain how non-competitive inhibitors work?

A

Non-competitive inhibitors bind not at AS but at an alternative site that maybe changes the structure of the AS, maybe changes the overall protein conformation a bit.

211
Q

Why do non-competitive inhibitors effect Vmax?

A

Non-competitive inhibitor binds at an alternative site and decreases the turnover number of the enzyme, thus lowering Vmax

212
Q

What is a significant thing to remember about non-competitive inhibitors?

A

Compounds binding outside the active site can also activate rather than inhibit.

We will see later on that there are things known as allosteric inhibitors or allosteric effectors that can work in a similar sort of way.

213
Q

Fig. 24

Label the image

A

Left: Substrate, Enzyme

Right: Substrate, Enzyme, Noncompetitive inhibitor

214
Q

Draw how a non-competitive inhibitor affects enzyme binding.

A

See Fig. 24

Left: Substrate, Enzyme
Right: Substrate, Enzyme, Noncompetitive inhibitor

Must show change in active site

215
Q

Km unaffected; Vmax decreases.

What does this describe?

A

Non-competitive inhibition.

216
Q

Why do non-competitive inhibitors (NCI) effect Vmax but not Km?

A

If we look at NCI then what you can see is actually Vmax drops bc you reach a point and whatever amount of substrate you add you cannot overcome inhibition, as the enzyme is binding away from the active site so the substrate concentration is unaffected (no competition, hence the name).

217
Q

How does a non-competitive inhibitor (NCI) appear on a Lineweaver-Burk (LwB) plot compared to an enzyme without an inhibitor?

A

See differences on intercept of y-axis but intercept of x-axis stays the same. So this tells us in NCI Vmax goes down and Km stays the same

218
Q

Fig. 35 (left)

Label the graph.

A
x-axis: [Substrate] -->
y-axis: Relative rate 0 20 40 60 80 100
Lines from top down:
Black: No inhibitor
[I] = Ki
[I] = 5 Ki
[I] = 10 Ki

Km for uninhibited enzyme

219
Q

Fig. 35 (right)

Label the graph.

A

x-axis: 1 / [S]
y-axis: 1 /V
black: No inhibitor present
pink: + Noncompetitive inhibitor

220
Q

Draw a Michaelis-Menten plot of enzyme activity with 3 non-competitive inhibitors (3 marks)

A

See Fig. 35 (left)

x-axis: [Substrate] -->
y-axis: Relative rate 0 20 40 60 80 100 (1 mark)
Lines from top down:
Black: No inhibitor
[I] = Ki
[I] = 5 Ki
[I] = 10 Ki

Km for uninhibited enzyme

Show Vmax affected (rate dropped) (1 mark)
Show Km is the same - relative rate has same concentration of Km. (1 mark)

221
Q

Draw a Lineweaver-Burk plot of enzyme activity with 3 non-competitive inhibitors (3 marks)

A

See Fig. 35 (right)

x-axis: 1 / [S]
y-axis: 1 /V (1 mark)
black: No inhibitor present
pink: + Noncompetitive inhibitor

Show x-axis intercept unchanged (1 mark) - same Km

Show y-axis intercept HIGHER with noncompetitive inhibitor (1 mark) - lower Vmax

222
Q

Summarise this lecture

A
  • enzymes work by lowering the activation energy of a reaction
  • the active site is the part of an enzyme where the reaction is catalysed
  • the active site only occupies a small proportion of the total enzyme and binds the substrate(s) via weak interactions
  • the rate of an enzyme catalysed reaction is related to the concentration of substrate and can be represented mathematically by the Michaelis-Menten equation
  • the kinetic parameters Km and Vmax provide information about the affinity of the enzyme for its substrate and the rate of reaction respectively
  • many compounds and drugs can interact with enzymes to inhibit the action of enzymes
223
Q

Summarise how enzymes work.

A

Enzymes work by lowering the activation energy of a reaction

224
Q

Summarise where reactions occur in enzymes.

A

The active site is the part of an enzyme where the reaction is catalysed

225
Q

Summarise the structure of the active site.

A

The active site only occupies a small proportion of the total enzyme and binds the substrate(s) via weak interactions

226
Q

Summarise the Michaelis-Menten equation.

A

The rate of an enzyme catalysed reaction is related to the concentration of substrate and can be represented mathematically by the Michaelis-Menten equation

227
Q

Summarise Km and Vmax.

A

The kinetic parameters Km and Vmax provide information about the affinity of the enzyme for its substrate and the rate of reaction respectively

228
Q

Summarise how drugs can be used as inhibitors.

A

Many compounds and drugs can interact with enzymes to inhibit the action of enzymes

229
Q

What do I need to know?

  1. Explain the effects of enzymes on chemical reactions
A
  • key features of enzymes

- role of the active site

230
Q

What do I need to know?

1a. Key features of enzymes

A

What they look like

231
Q

What do I need to know?

1b. Role of the active site

A

what are active sites

232
Q

What do I need to know?

  1. Describe how reaction rates vary as a function of enzyme and substrate concentration
A
  • simple graphs of velocity vs [S]

- how does the rate vary with enzyme concentration

233
Q

What do I need to know?

2a. simple graphs of velocity vs [S]

A

Could you plot this?

234
Q

What do I need to know?

2b. how does the rate vary with enzyme concentration

A

Work session question

235
Q

What do I need to know?

  1. Define the terms activity, international unit of enzyme activity, Km and Vmax
A

See cards

236
Q

What do I need to know?

3a. Define the term activity

A

See cards

237
Q

What do I need to know?

3b. Define the term international unit of enzyme activity

A

See cards

238
Q

What do I need to know?

3c. Define the term Km

A

See cards

239
Q

What do I need to know?

3d. Define the term Vmax

A

See cards

240
Q

What do I need to know?

  1. Analyse and interpret kinetic data for enzyme-catalysed reactions
A
  • ability to work out rates of enzyme-catalysed reactions

- know what Km and Vmax means to be able to apply this information

241
Q

What do I need to know?

4a. Ability to work out rates of enzyme-catalysed reactions

A

Understand graphs

242
Q

What do I need to know?

4b. Know what Km and Vmax means to be able to apply this information

A

Understand graphs

243
Q

What do I need to know?

  1. Describe the effects of enzyme inhibitors on enzyme kinetics and be able to distinguish between the two from simple graphs
A
  • why is a competitive inhibitor competitive?
  • show how an inhibitor would alter the shape of v vs [S] plot (or Lineweaver-Burk plot)
  • determine the type of inhibition from an of v vs [S] plot (or Lineweaver-Burk plot)
244
Q

What do I need to know?

5a. Why is a competitive inhibitor competitive?

A

Know the effects

245
Q

What do I need to know?

5b. Show how an inhibitor would alter the shape of v vs [S] plot (or Lineweaver-Burk plot)

A

Interpret graphs

246
Q

What do I need to know?

5c. Determine the type of inhibition from an of v vs [S] plot (or Lineweaver-Burk plot)

A

Interpret graphs and know effects of CI and NCI