Proteins, enzymes and enzyme kinetics

What you'll learn

How amino acids join to make proteins, and how protein structure links to function.
Why enzymes are specific biological catalysts with active sites.
How temperature, pH, substrate concentration and inhibitors affect enzyme rate.
How to calculate initial rate and interpret $VmaxV_\text{max}$ , $K_m$ and $Q_{10}$ .

1. Proteins are polymers of amino acids

Proteins are large biological molecules made from amino acids. There are 20 common amino acids used to make human proteins, and the order they are joined in affects the final shape and function of the protein.

Definition

Amino acid

An amino acid is a monomer with a central carbon atom bonded to an amino group, a carboxyl group, a hydrogen atom and a variable R group. The R group differs between amino acids and gives each one its chemical properties.

Amino acids join by a condensation reaction, where water is released and a peptide bond forms between the amino group of one amino acid and the carboxyl group of another. A chain of many amino acids is a polypeptide.

Proteins can be broken down by hydrolysis, where water is used to break peptide bonds.

2. Levels of protein structure

The function of a protein depends on its three-dimensional shape. That shape is built in levels:

Primary structure: the sequence of amino acids in the polypeptide.
Secondary structure: local folding into an alpha helix or beta pleated sheet, held by hydrogen bonds.
Tertiary structure: the overall three-dimensional shape of one polypeptide, held by interactions between R groups.
Quaternary structure: two or more polypeptide chains, or polypeptides plus non-protein groups, joined to form one functional protein.

This diagram summarises how amino acids join and how the levels of protein structure build up from sequence to final folded protein.

Protein structure from amino acids to quaternary structure

Key Idea

Sequence determines shape

A protein’s primary structure determines how it folds, because different R groups form different bonds and interactions. Shape then determines function.

Important interactions in tertiary structure include:

Hydrogen bonds between polar groups.
Ionic bonds between oppositely charged R groups.
Disulfide bridges between sulfur-containing R groups.
Hydrophobic interactions, where non-polar R groups cluster away from water.

Globular proteins, such as enzymes, are compact and usually soluble. Fibrous proteins, such as collagen, are long, strong and often structural.

Example

Predicting the effect of an amino acid substitution

A mutation changes one amino acid in an enzyme’s active site from a charged R group to a non-polar R group. Predict the likely effect.

The mutation changes the primary structure, because the amino acid sequence is altered.
Replacing a charged R group with a non-polar R group changes the possible bonding and charge distribution in that region of the protein.
If the active site shape or charge changes, the substrate may bind less easily, so fewer enzyme-substrate complexes form per second and the reaction rate decreases.

3. Enzymes are biological catalysts

An enzyme is a globular protein that increases the rate of a biochemical reaction without being used up.

Definition

Catalyst

A catalyst increases the rate of a reaction by lowering the activation energy, which is the minimum energy needed for a reaction to start.

The part of the enzyme that binds the substrate is the active site. The reacting molecule is the substrate. When the substrate binds, an enzyme-substrate complex forms.

Modern A-Level Biology usually explains enzyme specificity using the induced-fit model. The active site is not completely rigid; it changes shape slightly as the substrate binds. This helps strain bonds in the substrate and lowers activation energy.

The diagram shows the enzyme-substrate complex and the key enzyme kinetics graph you need to recognise.

Enzyme-substrate complex and Michaelis-Menten graph

Common Mistake

Lock-and-key is too simple

The lock-and-key model is useful for basic specificity, but exam answers are stronger if you describe induced fit: the active site changes shape slightly when the substrate binds.

4. Measuring enzyme rate

The rate of reaction is how quickly substrate is used up or product is formed.

In enzyme practicals, you usually calculate the initial rate. This is the rate at the start of the reaction, before substrate concentration has fallen much and before products have built up.

taken\text{rate} = \frac{\text{change in amount of product or substrate}}{\text{time taken}}

Example

Calculating an initial reaction rate

An enzyme produces 7.6 µmol of product in the first 40 s, and the graph is straight over this interval. Calculate the initial rate.

Use the straight initial section, because the gradient there best represents the initial rate before conditions change.
Substitute the values with units: $s−1\text{initial rate} = \frac{7.6\ \mu\text{mol} - 0.0\ \mu\text{mol}}{40\ \text{s} - 0\ \text{s}} = 0.190\ \mu\text{mol s}^{-1}$ .
State the result to a sensible precision: the initial rate is 0.19 µmol s⁻¹.

Tip

Using graphs

On a product-time graph, the rate is the gradient. If the curve bends, draw a tangent at time zero to estimate the initial rate.

5. Substrate concentration, $VmaxV_\text{max}$ and $K_m$

At low substrate concentration, enzyme active sites are mostly free. Increasing substrate concentration increases the chance of collisions, so rate rises.

At high substrate concentration, all active sites are occupied almost continuously. The enzyme is saturated, so adding more substrate no longer increases rate much.

Definition

Vmax and Km

$VmaxV_\text{max}$ is the maximum initial rate when enzyme active sites are saturated. $K_m$ is the substrate concentration at which the initial rate is half of $VmaxV_\text{max}$ .

A lower $K_m$ usually means the enzyme has a higher affinity for its substrate, because half-maximal rate is reached at a lower substrate concentration.

Example

Reading Km from a kinetics graph

A graph of initial rate against substrate concentration levels off at 80 µmol s⁻¹. The rate is 40 µmol s⁻¹ when substrate concentration is 2.5 mmol dm⁻³. Interpret these values.

Identify the plateau as $VmaxV_\text{max}$ , so $s−1V_\text{max} = 80\ \mu\text{mol s}^{-1}$ .
Half of this is $s−1\frac{80\ \mu\text{mol s}^{-1}}{2} = 40\ \mu\text{mol s}^{-1}$ .
The substrate concentration at this half-maximal rate is $K_m$ , so $dm−3K_m = 2.5\ \text{mmol dm}^{-3}$ .

6. Temperature and pH affect enzyme shape

Increasing temperature gives molecules more kinetic energy. This increases the frequency of successful collisions between enzyme and substrate, so rate usually increases up to an optimum temperature.

Above the optimum, bonds maintaining tertiary structure begin to break. The active site changes shape, so fewer enzyme-substrate complexes form. This is denaturation.

Definition

Denaturation

Denaturation is a change in a protein’s three-dimensional shape so that it can no longer function properly. In enzymes, this usually means the active site is no longer complementary to the substrate.

You may see temperature effects expressed using $Q_{10}$ , the factor by which rate changes for a 10 °C rise in temperature:

Q10=(R2R1)10T2−T1Q_{10} = \left(\frac{R_2}{R_1}\right)^{\frac{10}{T_2 - T_1}}

Example

Calculating Q10

An enzyme’s rate increases from 0.20 µmol s⁻¹ at 20 °C to 0.42 µmol s⁻¹ at 30 °C. Calculate $Q_{10}$ .

The temperature interval is 10 °C, so $10T2−T1=1030−20=1\frac{10}{T_2 - T_1} = \frac{10}{30 - 20} = 1$ .
Substitute the rates: $s−1)1=2.1Q_{10} = \left(\frac{0.42\ \mu\text{mol s}^{-1}}{0.20\ \mu\text{mol s}^{-1}}\right)^1 = 2.1$ .
$Q_{10}$ has no units, so the rate increased by a factor of 2.1 for a 10 °C rise.

Common Mistake

Q10 and denaturation

$Q_{10}$ is most useful below the denaturation range. Across or above the optimum temperature, rate may fall because enzyme structure is being disrupted.

pH also affects enzyme rate. Changing pH changes the concentration of hydrogen ions, which can alter ionic bonds and hydrogen bonds in the protein. Each enzyme has an optimum pH where its active site has the best shape and charge for binding the substrate.

7. Inhibitors reduce enzyme activity

An inhibitor is a substance that reduces enzyme activity.

A competitive inhibitor has a similar shape to the substrate and competes for the active site. Its effect can be reduced by increasing substrate concentration, because substrate molecules are more likely to occupy the active site.

A non-competitive inhibitor binds away from the active site and changes the enzyme’s shape or catalytic activity. Increasing substrate concentration does not fully overcome this because some enzyme molecules are made less effective.

Key Idea

Inhibitor effects on graphs

Competitive inhibition usually increases apparent $K_m$ but can still reach the same $VmaxV_\text{max}$ at very high substrate concentration. Non-competitive inhibition lowers $VmaxV_\text{max}$ .

8. Practical enzyme investigations

A common practical approach is to vary enzyme concentration, substrate concentration, temperature or pH, then measure initial rate.

For a valid investigation, control the other variables carefully:

Use a buffer to keep pH constant.
Use a water bath to control temperature.
Keep total volume constant.
Use the same enzyme and substrate concentrations unless they are the independent variable.
Repeat readings and calculate a mean.
Use a colorimeter or gas syringe where possible to reduce subjective judgement.

For example, catalase activity can be followed by measuring oxygen produced from hydrogen peroxide. Amylase activity can be followed by measuring starch breakdown using iodine, although judging the colour endpoint introduces more uncertainty.

Common Mistake

Using total product instead of initial rate

Do not compare enzyme activity using the final amount of product if reactions run for different times or reach completion. Compare initial rates for a fair kinetic comparison.

Exam technique

In the exam

When explaining rate changes, link the factor to active-site availability, collision frequency or denaturation.
For calculations, write the formula, substitute values with units, then state the final answer with units.
For enzyme graphs, identify the limiting factor: substrate concentration at low substrate levels, active-site saturation near $VmaxV_\text{max}$ .

Self review

Check yourself

Why can a change in primary structure alter the tertiary structure of a protein?
What does a low $K_m$ suggest about an enzyme-substrate pair?
How would you design a fair test of the effect of temperature on enzyme activity?

What you'll learn

How amino acids join to make proteins, and how protein structure links to function.
Why enzymes are specific biological catalysts with active sites.
How temperature, pH, substrate concentration and inhibitors affect enzyme rate.
How to calculate initial rate and interpret $VmaxV_\text{max}$ , $K_m$ and $Q_{10}$ .

1. Proteins are polymers of amino acids

Definition

Amino acid

Proteins can be broken down by hydrolysis, where water is used to break peptide bonds.

2. Levels of protein structure

The function of a protein depends on its three-dimensional shape. That shape is built in levels:

Primary structure: the sequence of amino acids in the polypeptide.
Secondary structure: local folding into an alpha helix or beta pleated sheet, held by hydrogen bonds.
Tertiary structure: the overall three-dimensional shape of one polypeptide, held by interactions between R groups.
Quaternary structure: two or more polypeptide chains, or polypeptides plus non-protein groups, joined to form one functional protein.

This diagram summarises how amino acids join and how the levels of protein structure build up from sequence to final folded protein.

Protein structure from amino acids to quaternary structure

Key Idea

Sequence determines shape

A protein’s primary structure determines how it folds, because different R groups form different bonds and interactions. Shape then determines function.

Important interactions in tertiary structure include:

Hydrogen bonds between polar groups.
Ionic bonds between oppositely charged R groups.
Disulfide bridges between sulfur-containing R groups.
Hydrophobic interactions, where non-polar R groups cluster away from water.

Globular proteins, such as enzymes, are compact and usually soluble. Fibrous proteins, such as collagen, are long, strong and often structural.

Example

Predicting the effect of an amino acid substitution

A mutation changes one amino acid in an enzyme’s active site from a charged R group to a non-polar R group. Predict the likely effect.

The mutation changes the primary structure, because the amino acid sequence is altered.
Replacing a charged R group with a non-polar R group changes the possible bonding and charge distribution in that region of the protein.
If the active site shape or charge changes, the substrate may bind less easily, so fewer enzyme-substrate complexes form per second and the reaction rate decreases.

3. Enzymes are biological catalysts

An enzyme is a globular protein that increases the rate of a biochemical reaction without being used up.

Definition

Catalyst

A catalyst increases the rate of a reaction by lowering the activation energy, which is the minimum energy needed for a reaction to start.

The part of the enzyme that binds the substrate is the active site. The reacting molecule is the substrate. When the substrate binds, an enzyme-substrate complex forms.

The diagram shows the enzyme-substrate complex and the key enzyme kinetics graph you need to recognise.

Enzyme-substrate complex and Michaelis-Menten graph

Common Mistake

Lock-and-key is too simple

The lock-and-key model is useful for basic specificity, but exam answers are stronger if you describe induced fit: the active site changes shape slightly when the substrate binds.

4. Measuring enzyme rate

The rate of reaction is how quickly substrate is used up or product is formed.

In enzyme practicals, you usually calculate the initial rate. This is the rate at the start of the reaction, before substrate concentration has fallen much and before products have built up.

taken\text{rate} = \frac{\text{change in amount of product or substrate}}{\text{time taken}}

Example

Calculating an initial reaction rate

An enzyme produces 7.6 µmol of product in the first 40 s, and the graph is straight over this interval. Calculate the initial rate.

Use the straight initial section, because the gradient there best represents the initial rate before conditions change.
Substitute the values with units: $s−1\text{initial rate} = \frac{7.6\ \mu\text{mol} - 0.0\ \mu\text{mol}}{40\ \text{s} - 0\ \text{s}} = 0.190\ \mu\text{mol s}^{-1}$ .
State the result to a sensible precision: the initial rate is 0.19 µmol s⁻¹.

Tip

Using graphs

On a product-time graph, the rate is the gradient. If the curve bends, draw a tangent at time zero to estimate the initial rate.

5. Substrate concentration, $VmaxV_\text{max}$ and $K_m$

At low substrate concentration, enzyme active sites are mostly free. Increasing substrate concentration increases the chance of collisions, so rate rises.

At high substrate concentration, all active sites are occupied almost continuously. The enzyme is saturated, so adding more substrate no longer increases rate much.

Definition

Vmax and Km

$VmaxV_\text{max}$ is the maximum initial rate when enzyme active sites are saturated. $K_m$ is the substrate concentration at which the initial rate is half of $VmaxV_\text{max}$ .

A lower $K_m$ usually means the enzyme has a higher affinity for its substrate, because half-maximal rate is reached at a lower substrate concentration.

Example

Reading Km from a kinetics graph

A graph of initial rate against substrate concentration levels off at 80 µmol s⁻¹. The rate is 40 µmol s⁻¹ when substrate concentration is 2.5 mmol dm⁻³. Interpret these values.

Identify the plateau as $VmaxV_\text{max}$ , so $s−1V_\text{max} = 80\ \mu\text{mol s}^{-1}$ .
Half of this is $s−1\frac{80\ \mu\text{mol s}^{-1}}{2} = 40\ \mu\text{mol s}^{-1}$ .
The substrate concentration at this half-maximal rate is $K_m$ , so $dm−3K_m = 2.5\ \text{mmol dm}^{-3}$ .

6. Temperature and pH affect enzyme shape

Increasing temperature gives molecules more kinetic energy. This increases the frequency of successful collisions between enzyme and substrate, so rate usually increases up to an optimum temperature.

Above the optimum, bonds maintaining tertiary structure begin to break. The active site changes shape, so fewer enzyme-substrate complexes form. This is denaturation.

Definition

Denaturation

You may see temperature effects expressed using $Q_{10}$ , the factor by which rate changes for a 10 °C rise in temperature:

Q10=(R2R1)10T2−T1Q_{10} = \left(\frac{R_2}{R_1}\right)^{\frac{10}{T_2 - T_1}}

Example

Calculating Q10

An enzyme’s rate increases from 0.20 µmol s⁻¹ at 20 °C to 0.42 µmol s⁻¹ at 30 °C. Calculate $Q_{10}$ .

The temperature interval is 10 °C, so $10T2−T1=1030−20=1\frac{10}{T_2 - T_1} = \frac{10}{30 - 20} = 1$ .
Substitute the rates: $s−1)1=2.1Q_{10} = \left(\frac{0.42\ \mu\text{mol s}^{-1}}{0.20\ \mu\text{mol s}^{-1}}\right)^1 = 2.1$ .
$Q_{10}$ has no units, so the rate increased by a factor of 2.1 for a 10 °C rise.

Common Mistake

Q10 and denaturation

$Q_{10}$ is most useful below the denaturation range. Across or above the optimum temperature, rate may fall because enzyme structure is being disrupted.

7. Inhibitors reduce enzyme activity

An inhibitor is a substance that reduces enzyme activity.

Key Idea

Inhibitor effects on graphs

Competitive inhibition usually increases apparent $K_m$ but can still reach the same $VmaxV_\text{max}$ at very high substrate concentration. Non-competitive inhibition lowers $VmaxV_\text{max}$ .

8. Practical enzyme investigations

A common practical approach is to vary enzyme concentration, substrate concentration, temperature or pH, then measure initial rate.

For a valid investigation, control the other variables carefully:

Use a buffer to keep pH constant.
Use a water bath to control temperature.
Keep total volume constant.
Use the same enzyme and substrate concentrations unless they are the independent variable.
Repeat readings and calculate a mean.
Use a colorimeter or gas syringe where possible to reduce subjective judgement.

Common Mistake

Using total product instead of initial rate

Do not compare enzyme activity using the final amount of product if reactions run for different times or reach completion. Compare initial rates for a fair kinetic comparison.

Exam technique

In the exam

When explaining rate changes, link the factor to active-site availability, collision frequency or denaturation.
For calculations, write the formula, substitute values with units, then state the final answer with units.
For enzyme graphs, identify the limiting factor: substrate concentration at low substrate levels, active-site saturation near $VmaxV_\text{max}$ .

Self review

Check yourself

Why can a change in primary structure alter the tertiary structure of a protein?
What does a low $K_m$ suggest about an enzyme-substrate pair?
How would you design a fair test of the effect of temperature on enzyme activity?

What you'll learn

1. Proteins are polymers of amino acids

2. Levels of protein structure

3. Enzymes are biological catalysts

4. Measuring enzyme rate

5. Substrate concentration, $VmaxV_\text{max}$ and $K_m$

6. Temperature and pH affect enzyme shape

7. Inhibitors reduce enzyme activity

8. Practical enzyme investigations

Review Proteins, enzymes and enzyme kinetics by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Proteins, enzymes and enzyme kinetics

What you'll learn

1. Proteins are polymers of amino acids

2. Levels of protein structure

3. Enzymes are biological catalysts

4. Measuring enzyme rate

5. Substrate concentration, $VmaxV_\text{max}$ and $K_m$

6. Temperature and pH affect enzyme shape

7. Inhibitors reduce enzyme activity

8. Practical enzyme investigations

Review Proteins, enzymes and enzyme kinetics by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Proteins, enzymes and enzyme kinetics

What you'll learn

1. Proteins are polymers of amino acids

2. Levels of protein structure

3. Enzymes are biological catalysts

4. Measuring enzyme rate

5. Substrate concentration, VmaxV_\text{max}Vmax​ and KmK_mKm​

6. Temperature and pH affect enzyme shape

7. Inhibitors reduce enzyme activity

8. Practical enzyme investigations

Review Proteins, enzymes and enzyme kinetics by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Proteins, enzymes and enzyme kinetics Revision Guide

Proteins, enzymes and enzyme kinetics

What you'll learn

1. Proteins are polymers of amino acids

2. Levels of protein structure

3. Enzymes are biological catalysts

4. Measuring enzyme rate

5. Substrate concentration, VmaxV_\text{max}Vmax​ and KmK_mKm​

6. Temperature and pH affect enzyme shape

7. Inhibitors reduce enzyme activity

8. Practical enzyme investigations

Review Proteins, enzymes and enzyme kinetics by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Proteins, enzymes and enzyme kinetics Revision Guide

5. Substrate concentration, $VmaxV_\text{max}$ and $K_m$

5. Substrate concentration, $VmaxV_\text{max}$ and $K_m$