Enzymes Flashcards
What are enzymes?
Biological catalysts that speed up chemical reactions
Substrate
A molecule that bonds to the active site of an enzyme to be catalysed by it in a reaction
Structure of an enzyme: active site
The tertiary structure of its protein(s) fold into the specific shape
Which determines the active site where substrate binds to because they are SIMILAR (not the same) in shape
Enzyme substrate complex
When the substrate is bonded to the active site of an enzyme forming this
How do enzyme substrate complexes form?
When the substrate and enzyme’s active site collide successfully
Because the active site is complementary to substrate after changing tertiary structure in induced fit model
So forms complex
Products
What is a product of the enzyme catalysed reaction eg proteins being broken down to amino acids
What is the job of enzymes?
To increase rate of a reaction
Because it’s a biological catalyst
How do enzymes increase rate of reaction?
When an enzyme-substrate complex is formed the substrate bonds are put under stress
Causing the activation energy of this catalysed reaction to lower (energy required to break bonds in substrate) so reaction is more likely to occur
What is the induced fit model?
Explains how an enzyme’s active site can bond to a substrate
Because they are complementary to each other however the active site has to mould itself
Description of the induced fit model in action
Substrate moves towards enzyme’ active site, and fits inside because they are SIMILAR in shape
The active site changes shape so it can be complementary to substrate and bind to form an enzyme-substrate complex (puts bonds under stress in substrate)
Why do only substrate molecules cause the enzyme’s active site to distort?
Because the bonds between amino acids in the active site of the enzyme and the substrate won’t be formed so the shape of the tertiary structure won’t change
What was the original theory for how enzymes are complementary to substrates?
The lock and key theory which said all enzymes active sites are already complementary to the substrates
A product time graph
Tells us the amount of product produced over the course of an enzyme catalysed reaction
Could be under certain conditions such as increased temperature etc
Description of a product/ substrate time graph for an enzyme catalysed reaction
Initially rapid because there is a high concentration of substrate molecules thus more enzyme substrate complexes
Decreasing rate of reaction in middle because no. Of substrate molecules is decreasing as more is made to products so lower rate of reaction (less enzyme substrate complexes formed)
Plateaus because no more substrate molecules collide with active site = stopped
How to find rate of reaction at a specific point on a product/substrate time graph?
Draw tangent at that point
What’s important to remember about finding the rate of reaction at a specific point using a tangent?
Can only be done on a product time graph
Because some graphs already have rate of reaction on the y axis so simply use the y axis
What factors affect enzyme catalysed reaction?
Temperature
Substrate/ enzyme concentration
pH
Competitive/ non competitive inhibitors
How to describe the factors effect on INCREASING enzymes rate of reaction?
This factors causes ____
Which increases frequency of successful collisions between enzyme’s active site and substrate in a haven time because ____
So number of enzyme-substrate complexes formed increase in a given time
Therefore more products are formed in a given time, increases rate of reaction
How to describe the factors effect on DECREASING enzymes rate of reaction?
This factor causes _____
Which decreases the frequency of successful collisions between enzyme and substrate in a given time because _____
So number of enzyme-substrate complexes formed decrease in a given time
So less products formed in a given time, decreasing rate of reaction
How to find a factor-rate graph?
Decide on specific factor eg pH
Select enzyme and measure amount of product formed over time
Draw graph for the results
Repeat for other pHs
Find the rates at the same time for each graph and draw a factor-rate graph
Temperature as a factor on a factor-rate graph description
Rate increases with the temperature
Before reaching an optimum
Then the rate decreases rapidly
Why does enzyme rate of reaction increase with increasing the temperature?
Increasing temp causes an increase in kinetic energy
Which increases frequency of successful collisions between enzyme’s active site and substrate because there is more rapid movement between molecules
So number of enzyme-substrate complexes formed increase
Therefore more products are formed in a given time, increases rate of reaction
Optimum temperature reached
When the temperature is at this point, it has achieved maximum frequency of collisions between active site and enzyme
So rate is at maximum peak = temperature is no longer limiting factor
Why does the rate of reaction drop rapidly after going beyond the optimum temperature?
Because at higher temps enzymes begin vibrating more rapidly so bonds in the tertiary structure end up breaking
Which changes the tertiary structure thus active site
So no longer is complementary to substrate and can’t catalyse reaction
Can the enzyme renature after it has denatured to increased temperature?
No because the tertiary structure had changed too much
Low pH
High concentration of H+ ions
High pH
Low concentration of H+ ions
Enzymes optimum pH
When the rate of reaction for an enzyme to catalyse is at its highest
Depends on the enzyme
pH-rate graph description
On the x axis, at the optimum pH there is a massive increase in rate of reaction
But before and after that pH the rate decreases very rapidly to 0
Why does rate of reaction decrease above and below optimum pH in enzyme catalyse reaction?
Because changing from the optimum disrupts tertiary structure of enzyme causing a change in shape of the active site
So less enzyme substrate complexes are formed because the active site no longer complementary to substrate
Sometimes extreme pH causes complete denaturation
How to find the pH of a solution given the concentration of hydrogen ions?
pH = -log [H+]
H+ = the concentration of hydrogen ions
How does increasing the substrate concentration (but keeping enzyme conc constant) increase rate of reaction?
Because at low concentrations not all active sites are occupied so conc is limiting factor
Increasing number of substrate molecules
Increases the frequency of successful collisions between substrate and active site increases
Increasing no. Of enzyme substrate complexes thus rate of reaction
Relationship between rate of reaction and substrate concentration
Directly proportional
Until it stops increasing and reaches its V max despite substrate concentration increasing
Enzyme’s v max
The maximum rate of reaction an enzyme can reach when saturated with substrate
Why does the enzyme reach its v max?
Because there are no more free active sites for the extra substrate molecules to collide with
So there’s no enough enzymes to form more enzyme-substrate complexes = rate of reaction stops increasing
How does increasing the enzyme concentration (keeping substrate concentration constant) increase rate of reaction?
Because there are more active sites available
And so increased successful collisions
More enzyme substrate complexes at any point
So increased rate
How do competitive inhibitors decrease the rate of reaction
The comp inhibitor binds to the active site
By occupying the active site the substrate cannot bind to the active site so less successful collisions
Less enzyme substrate complexes formed
So reduces rate of reaction
Competitive inhibitors
Molecules that bind to the enzymes active site which inhibits the reaction and reduces rate of reaction because they are similar in shape to substrate so prevent enzyme- substrate complexes from forming
Where do comp inhibitors bind to?
Active site
How to offset the effect of competitive inhibitors
Increase substrate concentration
So greater change it will occupy active site instead of inhibitors
Rate-substrate concentration with comp an inhibitor graph
Will reach the v max eventually than if there were no comp inhibitor
But the rate before is reduced compared to with no inhibitor
Can irreversible competitive inhibitors be reversed by increasing substrate concentration?
No because it permanently binds to active site so increasing substrate concentration won’t change as it can’t bind to active site
Non competitive inhibitors
Molecules that bind to the allosteric site of an enzyme which reduces rate of reaction
Where do non comp inhibitors bind to?
Allosteric site
How do non competitive inhibitors decrease rate of reaction?
Binds to allosteric site
Causes tertiary structure of enzyme to change
So active site changes shape = not complementary to substrate
So enzyme substrate complexes not formed decreasing rate of reaction
Can non competitive inhibitors be overcome by increasing substrate concentration?
No will never reach v max because active site changed shape
So even with more substrates they can’t bind to active site
When using enzymes in a practical what should the control be?
Boiled enzyme because then the enzyme denatures and thus cannot form enzyme substrate complexes
What happens when enzymes denature?
The bonds between r groups in tertiary structure are broken so tertiary structure is disrupted
Causing polypeptide chains to unfold and destroy active site
Substrate is no longer complementary to enzyme’s active site
Feed back inhibition loops
enzyme 1 –> enzyme 2 –> enzyme 3
If a product released in an enzyme catalysed reaction 3 inhibits enzyme 1 to turn off its own production
