1.4Biological Molecules
Proteins- Enzymes — AQA A Level Biology 7402 Revision Notes & Topical Questions
15 min read 26 parts Lesson 17 of 49 Unit quiz
Enzymes: Active Sites, Specificity and Induced Fit — AQA A LEVEL BIOLOGY 7402 notes: Explain how enzymes lower activation energy, describe enzyme specificity and the induced-fit model and outline the practical to measure activity.Many proteins are enzymes.
- enzyme-substrate complex
- induced-fit hypothesis
- Worked example 1
- Worked example 2
- Worked example 3
- Saying enzymes are used up in the reaction
Section 1 of 26Enzymes as biological catalysts
Many proteins are enzymes. They are 'biological' because they function in living systems and 'catalysts' because they speed up chemical reactions without being used up or changed. Critical to function is the active site, the region where the substrate binds.
Section 2 of 26Specificity and the enzyme-substrate complex
An enzyme has an active site with a specific shape that fits a specific substrate (the shapes are complementary), forming an enzyme-substrate complex.
Section 3 of 26Activation energy
Enzymes speed up reactions because they influence the stability of bonds in the reactants. Binding destabilises bonds in the substrate, making it more reactive. Enzymes lower the activation energy of a reaction by providing an alternative energy pathway, so more substrate molecules have enough energy to react at a given temperature.
Section 4 of 26Induced-fit model
In the induced-fit hypothesis the enzyme and its active site (and sometimes the substrate) change shape slightly as the substrate enters, giving a closer fit and stressing the substrate's bonds.
Section 5 of 26Required practical: measuring enzyme activity
The progress of an enzyme-catalysed reaction can be followed by measuring the rate of formation of a product (for example the volume of oxygen released by catalase from hydrogen peroxide, collected in a gas syringe or by displacing water), or the rate of disappearance of the substrate. Keep other variables constant and take repeats.
Section 6 of 26Enzymes and metabolism
Enzymes control metabolism by catalysing particular reactions in sequences called metabolic pathways.
- while others work outside cells (extracellular), such as digestive amylase.
- allowing organisms to function at temperatures that do not damage their cells.
Section 7 of 26Worked examples
Explain why an enzyme only catalyses one reaction. [3]3 marks
The active site has a specific tertiary structure whose shape is complementary to only one substrate, so only that substrate forms an enzyme-substrate complex.
Explain how an enzyme lowers activation energy. [2]2 marks
It binds the substrate and strains or destabilises its bonds, providing an alternative pathway with lower activation energy.
Describe the induced-fit model. [2]2 marks
The active site changes shape slightly when the substrate binds, so the fit becomes closer, forming the complex and weakening bonds.
Section 8 of 26Amylase and iodine
In this investigation the rate of substrate disappearance is used to compare rates of reaction under different conditions.
Section 9 of 26Using a colorimeter
A colorimeter measures light absorbance (how much light is absorbed) or transmission (how much passes through).
- so absorbance falls and transmission rises.
Section 10 of 26Variables and design
Enzyme rate experiments determine the effect of changing one particular variable on the reaction rate.
Section 11 of 26Drawing graphs and finding initial rate
Use appropriate linear scales on axes, plot the independent variable on the x-axis with units, plot points accurately and join with a line of best fit or smooth curve.
Section 12 of 26Common sources of error
In the amylase-iodine practical the end point is judged by eye, which is subjective; a colorimeter reduces this.
Section 13 of 26Worked examples
Describe how you would find the time for amylase to digest starch at different temperatures. [5]5 marks
Place starch and amylase in separate tubes in water baths at each temperature; mix; every 30 s remove a drop and add to iodine solution; record the time when the colour stays orange-brown. Keep pH (buffer), volumes and concentrations constant and repeat.
A tangent shows 72 cm3 of product formed in 20 s. Calculate the initial rate. [2]2 marks
72 ÷ 20 = 3.6 cm3 s-1.
Explain why a colorimeter is calibrated with weak iodine solution. [2]2 marks
It sets a reference (100% transmission) so readings correspond to the colour of the reaction mixture.
Section 14 of 26Temperature
Enzymes have an optimum temperature at which they catalyse at the maximum rate.
- so there is a lower frequency of successful collisions between substrate and active site.
- and the rate drops sharply.
Section 15 of 26pH
Each enzyme has an optimum pH.
- because the hydrogen and ionic bonds that hold the tertiary structure are broken by an excess of H+ (acidic) or OH- (alkaline) ions.
- whereas pH does not affect collision rate but changes in pH disrupt substrate binding,
- reducing successful collisions until the active site is no longer functional.
pH = −log10[H+]. Do not forget the minus sign; pH values should be between 0 and 14, and a weak acid should be about pH 4-6. For [H+] = 3.5 × 10-11 mol dm-3, pH = 10.456 (alkaline).
Section 16 of 26Enzyme concentration
The higher the enzyme concentration, the greater the number of active sites and the greater the likelihood of enzyme-substrate complexes forming. As long as there is enough substrate, the initial rate increases linearly with enzyme concentration. If substrate becomes limiting, the rate levels off.
Section 17 of 26Substrate concentration
The greater the substrate concentration, the higher the rate, because the chance of forming enzyme-substrate complexes increases.
- so the graph plateaus.
Section 18 of 26Worked examples
Explain the shape of the graph of rate against temperature for an enzyme. [5]5 marks
Rate rises with temperature because kinetic energy and collision frequency increase; maximum at the optimum; above it hydrogen and ionic bonds break, tertiary structure changes, active site loses its shape (denatured) and rate falls to zero.
Ethanoic acid from wood ants has [H+] = 8.39 × 10-6 mol dm-3. Find the pH. [2]2 marks
pH = −log(8.39 × 10-6) = 5.08.
Explain why the rate stops increasing at high substrate concentration. [3]3 marks
All active sites are occupied, so the enzyme works at Vmax; adding more substrate cannot increase the rate; enzyme concentration is limiting.
Section 19 of 26Reversible inhibitors
An enzyme's activity can be reduced or stopped temporarily by a reversible inhibitor. There are two types. Competitive inhibitors have a shape similar to the substrate and compete with it for the active site, so fewer enzyme-substrate complexes form. Non-competitive inhibitors bind to another site (an allosteric site) on the enzyme, changing the shape of the active site so the substrate can no longer bind.
Section 20 of 26End-product inhibition
Metabolic reactions can be controlled by using the end-product of a sequence of reactions as a non-competitive, reversible inhibitor.
Section 21 of 26Effect on the rate
| Inhibitor | Initial rate | Vmax | Effect of more substrate |
|---|---|---|---|
| Competitive | Reduced by blocking active sites | Still reached at high substrate concentration | Overcomes the inhibition |
| Non-competitive | Reduced | Reduced (fewer active enzymes) | Does not overcome inhibition |
Section 22 of 26Control of variables and uncertainty
In enzyme rate experiments only one factor (temperature, pH, enzyme, substrate or inhibitor concentration) is changed at a time.
- For example, with 50 cm3 of oxygen from the breakdown of hydrogen peroxide by catalase and an uncertainty of 0.5 cm3, the percentage error is 0.5 ÷ 50 × 100 = 1%.
Section 23 of 26Examples and exam application
A well-known competitive inhibitor is malonate, which resembles succinate and inhibits succinate dehydrogenase in respiration.
- and many drugs and poisons act as enzyme inhibitors.
Section 24 of 26Worked examples
Explain how a competitive inhibitor reduces the rate of reaction. [3]3 marks
It has a shape similar to the substrate and binds to the active site, so fewer substrate molecules can bind and fewer enzyme-substrate complexes form.
Why can high substrate concentration overcome competitive but not non-competitive inhibition? [3]3 marks
Extra substrate outcompetes the inhibitor for the active sites in competitive inhibition so Vmax is reached; non-competitive inhibitors permanently remove active enzymes while they are bound.
Calculate the percentage error for 50 cm3 of oxygen with an uncertainty of 0.5 cm3. [2]2 marks
(0.5 ÷ 50) × 100 = 1%.
Describe end-product inhibition and why it is useful. [3]3 marks
The end-product binds to another site, changing the active site and slowing the pathway; when levels fall it detaches. It prevents wasteful overproduction.
Section 25 of 26Key terms
| Term | Meaning |
|---|---|
| Enzyme | Biological catalyst, usually a protein |
| Active site | Region of an enzyme where the substrate binds |
| Enzyme-substrate complex | Temporary structure formed when substrate binds |
| Activation energy | Energy needed to start a reaction |
| Induced-fit | Active site adapts its shape to the substrate |
| Term | Meaning |
|---|---|
| Amylase | Enzyme that hydrolyses starch to maltose |
| Colorimeter | Instrument measuring light absorbed or transmitted |
| Initial rate | Rate at the very start, found from a tangent |
| Independent variable | Variable deliberately changed |
| Control variable | Variable kept constant |
| Term | Meaning |
|---|---|
| Optimum | Condition giving the maximum rate of reaction |
| Denaturation | Irreversible change of tertiary structure and active site shape |
| Buffer | Solution that maintains a constant pH |
| Vmax | Maximum rate when all active sites are saturated |
| Limiting factor | Factor in shortest supply that restricts the rate |
| Term | Meaning |
|---|---|
| Competitive inhibitor | Binds to the active site, competing with substrate |
| Non-competitive inhibitor | Binds elsewhere, altering the active site |
| End-product inhibition | Product inhibits an earlier enzyme in its pathway |
| Percentage error | (Uncertainty ÷ measured value) × 100 |
| Feedback loop | Output of a process regulates the process |
Section 26 of 26Exam technique
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