AS & A Level Biology · Syllabus 9700 · Section 3 of 19

Enzymes

Enzymes are globular proteins that catalyse the reactions of life by lowering activation energy. This topic covers the active site, the lock-and-key and induced-fit models, measuring rates with catalase, amylase and a colorimeter, the effects of temperature, pH and concentrations on rate, Vmax and Km, competitive and non-competitive inhibition, and immobilised enzymes.

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Syllabus coverage

  • State that enzymes are globular proteins that catalyse reactions inside cells (intracellular enzymes) or are secreted to catalyse reactions outside cells (extracellular enzymes)
  • Explain the mode of action of enzymes in terms of an active site, enzyme-substrate complex, lowering of activation energy and enzyme specificity, including the lock-and-key hypothesis and the induced-fit hypothesis
  • Investigate the progress of enzyme-catalysed reactions by measuring rates of formation of products using catalase and rates of disappearance of substrate using amylase
  • Outline the use of a colorimeter for measuring the progress of enzyme-catalysed reactions that involve colour changes
  • Investigate and explain the effects of temperature, pH (using buffer solutions), enzyme concentration, substrate concentration and inhibitor concentration on the rate of enzyme-catalysed reactions
  • Explain that the maximum rate of reaction (Vmax) is used to derive the Michaelis-Menten constant (Km), which is used to compare the affinity of different enzymes for their substrates
  • Explain the effects of reversible inhibitors, both competitive and non-competitive, on enzyme activity
  • Investigate the difference in activity between an enzyme immobilised in alginate and the same enzyme free in solution, and state the advantages of using immobilised enzymes

Part of the AS & A Level Biology syllabus 9700.

Enzyme action: lock and key, and induced fitAn enzyme is a globular protein with a groove called the active site. The substrate binds there and the reaction happens; the enzyme isunchanged and can be used again.Lock and key model1. Enzyme and substratesubstrateenzymeactive sitethe active site is complementaryto the substrate: only thissubstrate fits.2. Enzyme-substrate complex, then productsenzyme-substrate complexproductsThe substrate binds, the reaction happens and the productsare released, and the enzyme is unchanged.Induced fit model1. Enzyme and substratesubstrateenzymeactive sitethe active site is not anexact fit at first.2. The active site moulds around the substrateenzyme-substrate complexBonding of the substrate makes the active site change shapeso that it fits closely, straining the substrate bonds.
The lock-and-key model of enzyme action: a substrate molecule whose shape is complementary to the active site binds to form an enzyme-substrate complex, the bonds within the substrate are strained so that the activation energy is lowered, and the products are released leaving the enzyme unchanged and free to act again.

Enzymes are globular proteins, and they work inside cells and outside them

An enzyme is a globular protein that acts as a biological catalyst: it increases the rate of a reaction and comes out of that reaction unchanged. Globular proteins fold so that the hydrophobic R groups point inwards and the hydrophilic R groups point outwards, which makes them compact and soluble, and the folding brings a small number of amino acid R groups together on the surface into a pocket called the active site. That pocket is where the chemistry happens. A single enzyme molecule can catalyse many thousands of reactions every second, so only a very small mass of enzyme is needed to convert a large mass of substrate.

Intracellular enzymes stay inside the cell. Catalase, for example, is held in peroxisomes in liver and potato cells, where it breaks down hydrogen peroxide, a toxic by-product of metabolism, into water and oxygen. DNA polymerase works in the nucleus, the enzymes of glycolysis work in the cytoplasm, and the enzymes of the Krebs cycle are in the mitochondrial matrix. Keeping an enzyme inside a membrane-bound compartment also keeps its reactions away from the rest of the cell, so a protein-digesting enzyme stored in a lysosome cannot digest the cell that made it.

Extracellular enzymes are secreted from the cell and catalyse reactions outside it. The digestive enzymes of mammals are the classic example: the salivary glands secrete amylase into the mouth, the pancreas secretes amylase, trypsin and lipase into the small intestine, and the glands in the stomach lining secrete pepsin into the stomach lumen. Secretion means packaging the enzyme into a vesicle that travels to the cell surface membrane and fuses with it. Because an extracellular enzyme works in the fluid of a cavity, its optimum pH matches that fluid: pepsin has an optimum near pH 2, matching the hydrochloric acid of the stomach, while trypsin has an optimum near pH 8 in the alkaline contents of the duodenum.

Cells that digest their surroundings depend on extracellular enzymes too. Saprotrophic fungi such as Rhizopus and the wood-rotting fungus Trametes versicolor secrete amylases, proteases and cellulases onto dead material, digesting it outside the hyphae and absorbing the soluble products. The same principle is exploited by industry, which harvests enzymes from microorganisms and uses them in biological washing powders, in cheese-making and in producing glucose syrup. The DNA polymerase of the thermophilic bacterium Thermus aquaticus, which lives in the hot springs of Yellowstone National Park, is the source of the heat-stable enzyme used in the polymerase chain reaction.

  • Intracellular: catalase in peroxisomes breaks down hydrogen peroxide; DNA polymerase in the nucleus; Krebs cycle enzymes in the mitochondrial matrix.
  • Extracellular: salivary amylase into the mouth, pancreatic amylase, trypsin and lipase into the small intestine, pepsin into the stomach.
  • Extracellular enzymes are secreted in vesicles and have optima that match the fluid they work in: pepsin about pH 2, trypsin about pH 8.
  • Enzymes are not used up. One molecule catalyses the same reaction many thousands of times before it is broken down.
Every enzyme is a globular protein with an active site. What makes an enzyme intracellular or extracellular is where the cell sends it, not what it is made of.

The active site, the enzyme-substrate complex and activation energy

The active site is a small region of the enzyme surface, built from only a few of the amino acids in the chain, that is complementary in shape and charge to its substrate. Substrate molecules bind there by forming temporary hydrogen bonds and ionic bonds with the R groups lining the pocket, producing an enzyme-substrate complex. While the substrate is held in place the reaction takes place, the products are released, and the enzyme is left exactly as it was and free to bind the next substrate molecule.

Activation energy is the minimum energy that the reacting molecules must have for the reaction to happen. At body temperature only a tiny fraction of molecules have anything like that much energy, so an uncatalysed reaction would be far too slow to keep a cell alive. The active site holds the substrate in the correct orientation and puts a strain on its bonds, offering an alternative route with a lower activation energy. Many more molecules now have enough energy to react, so the rate rises enormously. The enzyme does not change the overall energy change of the reaction and it does not change the equilibrium, only the speed with which equilibrium is reached.

Specificity means an enzyme catalyses one reaction, or a small group of closely related reactions, and it comes from the shape and the charge of the active site. The lock-and-key hypothesis, proposed by Emil Fischer in 1894, treats the active site as a rigid shape, the lock, into which only a substrate with a complementary shape, the key, fits. A substrate of a different shape cannot bind, no enzyme-substrate complex forms and no reaction is catalysed. This explains why amylase acts on starch but not on a protein, but it cannot explain why the active site of a real enzyme is not in fact rigid.

The induced-fit hypothesis, put forward by Daniel Koshland in 1958, says that the active site is flexible. When the substrate approaches, the enzyme changes shape slightly so that the active site moulds around the substrate and the fit becomes closer. The change of shape brings the reactive groups of the active site into the right position and puts strain on the bonds within the substrate, so the activation energy is lowered even further. X-ray crystallography of enzymes with and without their substrate bound shows these conformational changes directly, and induced fit also explains why some enzymes act on a family of similar substrates rather than on one molecule only.

FeatureLock-and-key hypothesisInduced-fit hypothesis
Shape of the active siteRigid, with a fixed shapeFlexible, and it changes shape when the substrate binds
How the substrate fitsThe substrate is a precise fit, like a key in a lockThe active site moulds around the substrate so that the fit becomes closer
What it explainsEnzyme specificity, and why a substrate of the wrong shape cannot bindHow activation energy is lowered by straining bonds, and why some enzymes act on a family of similar substrates
Proposed byEmil Fischer, 1894Daniel Koshland, 1958
The active site is complementary to the substrate, the enzyme lowers the activation energy, and the enzyme is unchanged at the end of the reaction.

Following the progress of a reaction: catalase, amylase and the colorimeter

To follow an enzyme-catalysed reaction you either measure the appearance of a product or the disappearance of the substrate, and you take readings at fixed time intervals rather than only at the end. The formation of a product is easy to follow with catalase, which catalyses the breakdown of hydrogen peroxide: 2H2O2 -> 2H2O + O2. Catalase from liver or potato tissue is extremely fast, so use a fixed small mass of tissue in a fixed volume of buffer, add a fixed volume of hydrogen peroxide solution, and collect the oxygen in a gas syringe or over water in an inverted burette, reading the volume every 10 or 20 seconds. The speed of bubbling may be used as a rough guide, but the volume of gas is far better because it is quantitative and gives numbers you can plot.

Disappearance of the substrate is measured with amylase, which hydrolyses starch to maltose. Take a sample from the mixture at fixed time intervals and test each one with iodine solution: while starch is present the iodine turns blue-black, and the end point, called the achromatic point, is the first time the iodine stays yellow-brown, showing that all the starch has gone. The rate is then found as 1 / time, because the less time the reaction takes, the faster it is. This method is quick but crude: the end point depends on your eye judging a colour, a sample is used up at every test, and only one value per tube is produced.

A colorimeter measures how much light of a chosen wavelength is absorbed by a coloured solution, so it follows any reaction in which the colour changes and turns the iodine end point into a continuous, quantitative measurement. Set the instrument to a filter of the colour complementary to the solution, because that is the light the solution absorbs most strongly: the blue-black starch-iodine complex absorbs red light, so a red filter is used. Calibrate the colorimeter with a blank made of buffer, iodine and distilled water but no starch, then take a reading immediately after the enzyme is added and further readings at fixed intervals. Absorbance falls as the starch disappears.

Plot a graph of absorbance against time. The reaction is fastest at the start, so the curve is steepest then, and the rate of reaction is the gradient of the initial straight part of the curve, before the graph begins to flatten. Taking a single absorbance reading and calling it the rate scores nothing: absorbance measures the concentration of the coloured substance, and a rate is a change in concentration divided by the time taken. The colorimeter is more reliable than an iodine end point judged by eye because it removes personal judgement of colour, it detects small changes in the amount of starch, and it produces enough data to show whether the reaction follows the expected pattern.

  • Product formation: catalase on hydrogen peroxide, with the oxygen collected in a gas syringe and the volume read every 10 or 20 seconds.
  • Substrate disappearance: amylase on starch, followed with iodine solution to the achromatic point, rate = 1 / time.
  • Colorimeter: absorbance of a chosen wavelength, red filter for the blue-black starch-iodine complex, calibrated with a blank.
  • Rate is the gradient of the initial straight part of the curve, not a single reading and not the final plateau.
  • Control everything except the variable you are testing: volumes, concentrations, temperature and pH, with a buffer holding the pH steady.
Measure the appearance of the product or the disappearance of the substrate, and always take the rate from the initial part of the curve, where the reaction is fastest.

Temperature and pH: the two factors that alter the enzyme itself

As the temperature rises, the rate of an enzyme-catalysed reaction increases, because enzyme and substrate molecules gain kinetic energy and collide more often and with more energy. More enzyme-substrate complexes form each second and more product is made. Up to the optimum temperature the effect is roughly exponential: for most enzymes working between 0 and 40 °C the rate about doubles for every 10 °C rise.

Above the optimum the rate falls sharply, and the fall is steep because it is caused by denaturation. The extra kinetic energy breaks the hydrogen bonds and ionic bonds that hold the tertiary structure of the globular protein in shape, the carefully folded chain unfolds, and the active site loses its shape, so it is no longer complementary to the substrate. Fewer enzyme-substrate complexes form and the rate collapses. Denaturation is a permanent change to the tertiary structure: cooling the mixture does not bring the activity back. This is why an enzyme below its optimum is inactivated but not denatured - at a low temperature the reaction is simply very slow because the molecules have little kinetic energy - whereas an enzyme heated above its optimum is destroyed as a catalyst.

The optimum temperature is not a universal constant: it reflects the temperature at which the organism normally lives. Human enzymes have an optimum near 37 °C, enzymes of the thermophilic bacterium Thermus aquaticus keep working above 70 °C, and enzymes of fish living in polar seas may have optima below 10 °C. Because their tertiary structure is held together by more or stronger interactions between R groups, thermostable enzymes survive conditions that would denature an ordinary enzyme, which is why they are used in industrial processes. In practical work the temperature is controlled with a water bath, and the enzyme and substrate solutions are left in it long enough to reach that temperature before they are mixed.

pH changes enzyme activity because hydrogen ions and hydroxide ions alter the charges on the R groups of the amino acids, both in the active site and elsewhere in the molecule. Away from the optimum pH, the hydrogen bonds and ionic bonds holding the tertiary structure break, the active site changes shape and the enzyme is denatured. Each enzyme has an optimum pH at which its activity is highest and at which the charges in the active site allow the substrate to bind and the reaction to be catalysed: pepsin works best at about pH 2 in the stomach, salivary amylase at about pH 6.8 in the mouth, and trypsin at about pH 8 in the alkaline small intestine. Extremes of pH in either direction denature an enzyme irreversibly. In investigations a buffer solution holds the pH steady: a buffer is a mixture, such as a weak acid with its sodium salt, that resists a change in pH when a small amount of acid or alkali is added, and it is used either to keep pH constant while another variable is tested or to set up a series of tubes at a range of pH values.

  • Rising temperature: more kinetic energy, more frequent and more energetic collisions, more enzyme-substrate complexes per second.
  • Above the optimum: the hydrogen bonds and ionic bonds of the tertiary structure break, the active site loses its shape, and the enzyme is denatured permanently.
  • Low temperature: the enzyme is inactivated, not denatured. The active site is intact and activity returns on warming.
  • Optimum temperature matches the habitat: about 37 °C for human enzymes, above 70 °C for enzymes of Thermus aquaticus.
  • pH alters the charges on R groups and breaks the bonds holding the tertiary structure; every enzyme has an optimum pH.
  • Use a buffer to hold pH constant, or to create a series of pH values, when pH is the independent variable.
Heat above the optimum denatures an enzyme permanently; cold only slows it down. A denatured enzyme has lost the shape of its active site and cannot be restored.

Enzyme concentration, substrate concentration, Vmax and Km

When substrate is in plentiful supply, the rate of reaction is directly proportional to the enzyme concentration: double the enzyme and you double the rate. The number of active sites available is limiting the reaction, so more enzyme molecules means more enzyme-substrate complexes can form each second. The straight-line relationship holds only while substrate is in excess. If the substrate is nearly used up, adding more enzyme achieves nothing, and the graph flattens because substrate has become the limiting factor.

The effect of substrate concentration follows the shape of a rectangular hyperbola rather than a straight line. At low substrate concentrations the rate is approximately proportional to the substrate concentration, because substrate molecules are scarce and the chance of one colliding with an active site rises as you add more. As the concentration is increased further the rise in rate slows down, and eventually the curve flattens into a plateau. At that point every active site is occupied nearly all the time, so the enzyme is saturated and adding more substrate has no effect: enzyme concentration is now the limiting factor, and the plateau is the maximum rate of reaction, Vmax.

Vmax is therefore the maximum rate of reaction, reached when every active site is occupied at all times. From the graph of rate against substrate concentration you can derive the Michaelis-Menten constant, Km, which is defined as the substrate concentration at which the rate of reaction is half of Vmax. Km has the units of concentration, such as mmol/dm3. Its value is used to compare the affinity of different enzymes for their substrates. An enzyme with a low Km reaches half of its maximum rate at a low substrate concentration, so it binds its substrate tightly, has a high affinity for it and works well even when substrate is scarce. An enzyme with a high Km needs a large substrate concentration to reach half Vmax, so its affinity is lower. Km is constant for a given enzyme under fixed conditions of temperature and pH, so two enzymes can only be compared fairly if they were measured under the same conditions.

The plateau is never quite reached, so Vmax cannot be read directly off the curve; it is estimated. One way is to plot the reciprocal values: a graph of 1 / rate against 1 / substrate concentration gives a straight line called a double-reciprocal or Lineweaver-Burk plot, for which the intercept on the vertical axis is 1 / Vmax and the intercept on the horizontal axis is -1 / Km. The relationship between the three quantities is rate = (Vmax x [S]) / (Km + [S]), where [S] is the substrate concentration, and putting [S] = Km into that expression gives rate = Vmax / 2, which is exactly what the definition of Km says. A real comparison is the pair of enzymes that phosphorylate glucose: hexokinase has a much lower Km for glucose than glucokinase, so hexokinase keeps working at the low glucose concentrations found in most cells, while glucokinase in liver cells only becomes fully active after a meal, when the glucose concentration is high.

  • Enzyme concentration: the rate is directly proportional, provided that substrate is in excess.
  • Substrate concentration: proportional at low concentrations, then a plateau when the active sites are saturated.
  • Vmax: the maximum rate of reaction, reached when every active site is occupied at all times.
  • Km: the substrate concentration at which the rate is half of Vmax, measured in units of concentration.
  • A low Km means a high affinity for the substrate; a high Km means a low affinity.
  • Km is constant for an enzyme only at fixed temperature and pH, so comparisons must be made under the same conditions.
Saturation explains the plateau. At high substrate concentration every active site is busy nearly all the time, so the enzyme concentration sets the rate and it cannot go any faster.
The effect of temperature on the rate of an enzyme-controlled reactionThe rate rises to an optimum and then falls sharply as the enzyme is denatured.0102030405060020406080100temperature / degrees Crate of reactionoptimum (about 40 degrees C)As the temperature rises the molecules gainkinetic energy, collide more often and formmore enzyme-substrate complexes, so the raterises.Above the optimum the enzymedenatures: the hydrogen bondsthat hold the active site inshape break, the substrate nolonger fits and the rate falls.The fall after the optimum is not the reverse of the rise: denaturation is permanent, so cooling themixture does not restore the rate.
Graph of rate of reaction against substrate concentration for an enzyme-catalysed reaction: the rate rises almost in proportion at low substrate concentration, then levels off at the maximum rate Vmax as the active sites become saturated; the substrate concentration at which the rate is half of Vmax is the Michaelis-Menten constant, Km.

Reversible inhibition: competitive and non-competitive

An inhibitor is a substance that slows an enzyme-catalysed reaction. Reversible inhibitors bind to the enzyme by weak hydrogen bonds and ionic bonds, so they can leave again and the enzyme returns to full activity. This makes them different from irreversible inhibitors such as the heavy metal ions mercury and silver, which bind permanently to sulphydryl groups, and cyanide, which binds irreversibly to the iron in its target enzyme.

A competitive inhibitor has a shape similar to the substrate, so it can enter the active site and compete with the substrate for it. While the inhibitor occupies the active site an enzyme-inhibitor complex forms and no product is made. Because both molecules are competing for the same site, the outcome depends on the relative concentrations: raise the substrate concentration high enough and the substrate wins, the inhibition is overcome and the reaction still reaches the original Vmax. The apparent Km rises, however, because a higher substrate concentration is now needed to reach half of the maximum rate, and a raised Km is exactly what a reduced affinity between enzyme and substrate looks like. Malonate is the classic example: it is similar in shape to succinate and competes with it for the active site of succinate dehydrogenase, so the Krebs cycle slows down. The same competition is used in medicine, where a patient who has swallowed methanol is given ethanol, because ethanol competes with methanol for the active site of alcohol dehydrogenase and so reduces the amount of methanol converted into the harmful methanoic acid.

A non-competitive inhibitor binds to the enzyme at a site away from the active site. It bears no resemblance to the substrate and does not compete with it. Binding at this second site changes the shape of the enzyme molecule, and because the tertiary structure is distorted the active site changes shape as well, so the substrate is no longer complementary to it and enzyme-substrate complexes form less readily or break down more slowly. Adding more substrate does not help, because the substrate is not being blocked at the active site: the enzyme itself has been altered. As a result Vmax is reduced, since less working enzyme is available and the plateau of the graph is lower, while Km is unchanged, because the enzyme molecules that remain active still bind the substrate with the same affinity. Cyanide is a non-competitive inhibitor of cytochrome oxidase, the final enzyme of the electron transport chain: it binds to the iron in the enzyme and stops the chain, so oxidative phosphorylation and ATP synthesis stop, which is why cyanide is so rapidly fatal. End-product inhibition works the same way, as when ATP binds at an allosteric site on phosphofructokinase and slows glycolysis.

You tell the two apart from a graph of rate against substrate concentration. With a competitive inhibitor the curve still reaches the same Vmax but a higher substrate concentration is needed to get there, so Km appears higher. With a non-competitive inhibitor the curve levels off at a lower Vmax while the value of Km is unchanged. The size of the effect of either kind of inhibitor depends on its concentration, and in practical work the enzyme and substrate volumes and concentrations, the pH buffer and the temperature must all be held constant while the inhibitor concentration is changed.

FeatureCompetitive inhibitorNon-competitive inhibitor
Where it bindsThe active siteA site away from the active site
Similarity to the substrateSimilar shape to the substrate, so it fits the active siteNo resemblance to the substrate
Effect of raising substrate concentrationOvercomes the inhibition, and Vmax is still reachedDoes not overcome the inhibition
Effect on VmaxUnchangedReduced, because less active enzyme is available
Effect on KmApparent increase, because more substrate is needed to reach half VmaxUnchanged, because affinity for the substrate is unaltered
ExampleMalonate competing with succinate at succinate dehydrogenase; ethanol competing with methanol at alcohol dehydrogenaseCyanide binding to cytochrome oxidase; ATP inhibiting phosphofructokinase as end-product inhibition
Competitive: same Vmax, higher Km. Non-competitive: lower Vmax, same Km. The inhibitor that can be out-competed raises Km; the one that cannot lowers Vmax.

Immobilised enzymes in calcium alginate, and why industry prefers them

An immobilised enzyme is one that has been trapped inside, or attached to, an insoluble support, so that it stays in place while the substrate flows past it. The method used in school is entrapment in calcium alginate. Sodium alginate, a polysaccharide extracted from brown seaweed, is dissolved in water together with the enzyme, and the mixture is dripped from a syringe into calcium chloride solution. Calcium ions displace sodium ions and the alginate sets into insoluble beads of calcium alginate gel, with the enzyme molecules trapped inside the network of the gel. The beads are then packed into a column or held in a reaction vessel and the substrate solution is passed through or over them.

To compare an immobilised enzyme with the same enzyme free in solution, use the same mass of enzyme in each case - one portion in the beads and one in solution - with the same substrate concentration, the same buffer, the same temperature and the same total volume, and measure either the time taken to break down a fixed amount of substrate or the amount of product formed after a fixed time. The free enzyme is usually faster, and the reasons are structural rather than chemical. Trapped enzyme molecules cannot move freely to meet the substrate, so collisions are less frequent; the substrate must diffuse into the gel to reach the active sites, and that diffusion is slow compared with diffusion in a stirred solution; some enzyme molecules sit deep inside a bead where their active sites are partly blocked by the alginate; and a column gives results that change with the flow rate for the same reason.

The advantages of immobilisation are what make the loss of speed worth accepting. The enzyme is easy to separate from the product: the reaction mixture is simply drained off, or the beads are filtered out, so the product is not contaminated with enzyme and no step is needed to destroy or remove the enzyme afterwards. The enzyme can be reused many times, which is a large saving because enzymes are expensive to produce and purify, and the process can be run continuously in a packed column instead of in separate batches. An immobilised enzyme is also more stable to heat and to changes in pH than the same enzyme in solution, because the support holds the tertiary structure in place, so the reaction can be run at a higher temperature and therefore faster. Because it is held on a support, the enzyme is also not washed away in continuous flow.

Industry uses immobilised enzymes on a large scale. Immobilised lactase is used to make lactose-free milk and to produce a glucose and galactose syrup for ice cream and confectionery, and because the enzyme is held in a column it can be reused for weeks. Immobilised glucose isomerase converts glucose syrup into the sweeter high-fructose syrup used in soft drinks. Immobilised penicillin acylase (penicillin amidase) hydrolyses natural penicillin to 6-aminopenicillanic acid, the starting material for the semisynthetic penicillins, and immobilised aminoacylase is used to resolve a mixture of amino acids so that only the L-form needed for food and pharmaceutical use is produced.

  • Immobilised enzyme: an enzyme held inside or on an insoluble support, such as beads of calcium alginate gel.
  • Making the beads: mix the enzyme with sodium alginate solution, then drip the mixture into calcium chloride solution so the beads set.
  • The free enzyme is usually faster, because trapped enzyme molecules cannot move freely and the substrate must diffuse into the gel.
  • Advantages: easy separation from the product, reuse of an expensive catalyst, continuous operation, and greater stability to heat and pH.
  • Uses: lactase for lactose-free milk, glucose isomerase for high-fructose syrup, penicillin acylase for semisynthetic penicillins.
Immobilisation does not make an enzyme work faster. It makes the enzyme reusable, stable and easy to separate from the product, which is what matters in an industrial process.

Key terms you must be able to define

Enzyme
A globular protein that acts as a biological catalyst: it increases the rate of a reaction by lowering its activation energy and is unchanged at the end of the reaction.
Active site
The small region of the enzyme surface, formed by a few amino acid R groups, that is complementary in shape and charge to the substrate and where the reaction is catalysed.
Enzyme-substrate complex
The temporary complex formed when substrate molecules bind in the active site, held by hydrogen bonds and ionic bonds between the substrate and the R groups lining the active site.
Activation energy
The minimum energy that the reacting molecules must have for a reaction to take place. Enzymes lower it by holding the substrate in the correct orientation and straining its bonds, so many more molecules can react and the rate increases.
Denaturation
The permanent loss of the tertiary structure of an enzyme, caused by a temperature above the optimum or by extreme pH, which changes the shape of the active site so that it is no longer complementary to the substrate.
Vmax
The maximum rate of reaction of an enzyme-catalysed reaction, reached when every active site is occupied by substrate so that the active sites are saturated and enzyme concentration is the limiting factor.
Michaelis-Menten constant (Km)
The substrate concentration at which the rate of reaction is half of Vmax; it is used to compare the affinity of different enzymes for their substrates, a low Km meaning binding is tight and the affinity is high. It is constant only at fixed temperature and pH.
Competitive inhibitor
A molecule with a shape similar to the substrate that binds to the active site and competes with the substrate for it; its effect is overcome by raising the substrate concentration, so Vmax is unchanged but the apparent Km rises.
Non-competitive inhibitor
A molecule that binds to the enzyme at a site away from the active site and changes the shape of the enzyme, so that the active site is no longer complementary to the substrate; more substrate does not overcome it, so Vmax is reduced while Km is unchanged.

Applied to the real world

Cambridge sets 9700 in unfamiliar contexts - a hospital, a research laboratory, a farm, a nature reserve. Use these as models for how to apply the biology, then write your own.

In a food factory

Lactase immobilised in calcium alginate beads is used to make lactose-free milk. The milk is pumped through a column packed with the beads, and the lactase hydrolyses the lactose into glucose and galactose, which are sweeter and easier to digest. Because the enzyme is trapped in the beads it never leaves the column, so the milk is not contaminated with enzyme and the same batch of lactase can be used for weeks. The immobilised enzyme is also more stable, so the process can be run at a temperature that would quickly denature the free enzyme.

In a laboratory

A student measures the activity of catalase by dropping a fixed small piece of potato into a fixed volume of hydrogen peroxide solution buffered to pH 7 and held at 25 °C in a water bath, collecting the oxygen in a gas syringe and reading the volume every 20 seconds. The initial rate is the gradient of the steepest part of the curve. The same reaction can be followed with amylase and starch instead, taking samples at fixed intervals and using a colorimeter with a red filter, because the blue-black starch-iodine complex absorbs red light.

In a hospital

A patient who has swallowed methanol is treated by giving ethanol, an application of competitive inhibition. Alcohol dehydrogenase in the liver acts on both alcohols, but it oxidises methanol to methanoic acid, which damages the optic nerve and the kidneys. Ethanol occupies the active site of the enzyme for much of the time, so less methanol is metabolised and far less methanoic acid is formed, giving the kidneys time to excrete the methanol. The inhibition is competitive because raising the concentration of one alcohol out-competes the other for the same active site.

On a farm

Urea spread as a fertiliser is broken down by the extracellular enzyme urease, released by soil bacteria and by plant roots. The ammonia produced can escape into the air as a gas, which wastes nitrogen and adds to air pollution. Farmers reduce the loss by adding a urease inhibitor such as NBPT to the urea, so that the enzyme is inhibited, the ammonia is released more slowly and more of the nitrogen stays in the soil where the crop roots can take it up.

In a nature reserve

The hot springs of Yellowstone National Park are home to the thermophilic bacterium Thermus aquaticus, whose enzymes have optima above 70 °C and are not denatured even near the temperature of boiling water. Its DNA polymerase, known as Taq polymerase, survives the repeated heating steps of the polymerase chain reaction, so it can copy DNA again and again without being replaced. The enzyme is a direct product of natural selection for a tertiary structure held together by unusually strong interactions between its R groups.

Flashcards

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Q1What is an enzyme?
A globular protein that acts as a biological catalyst: it increases the rate of a reaction by lowering its activation energy, and it is unchanged at the end of the reaction.
Q2What is the difference between an intracellular and an extracellular enzyme?
Intracellular enzymes catalyse reactions inside the cell and stay there, such as catalase in peroxisomes and DNA polymerase in the nucleus. Extracellular enzymes are secreted in vesicles and catalyse reactions outside the cell, such as salivary amylase, pepsin and trypsin in the gut lumen.
Q3What is an active site?
A small region of the enzyme surface, formed by a few amino acid R groups, which is complementary in shape and charge to the substrate and where the reaction is catalysed.
Q4How does the lock-and-key hypothesis differ from the induced-fit hypothesis?
Lock-and-key (Fischer, 1894): the active site is rigid and the substrate fits it exactly like a key in a lock. Induced fit (Koshland, 1958): the active site is flexible and changes shape as the substrate binds, moulding around it and straining its bonds.
Q5What is an enzyme-substrate complex, and what holds it together?
The temporary complex formed when substrate molecules bind in the active site, held by hydrogen bonds and ionic bonds between the substrate and the R groups of the active site.
Q6How does an enzyme increase the rate of a reaction?
It lowers the activation energy by holding the substrate in the correct orientation and straining its bonds, so a much larger proportion of molecules have enough energy to react.
Q7Why does enzyme activity fall above the optimum temperature?
The extra kinetic energy breaks the hydrogen bonds and ionic bonds holding the tertiary structure, so the enzyme unfolds, the active site loses its shape, few enzyme-substrate complexes form and the enzyme is denatured permanently.
Q8Why does pH change the activity of an enzyme?
Hydrogen and hydroxide ions change the charges on the R groups in the molecule, which breaks the hydrogen and ionic bonds holding the tertiary structure, so the active site loses its shape and the enzyme is denatured. Each enzyme has an optimum pH.
Q9What happens to the rate as substrate concentration is increased?
The rate rises almost in proportion at low substrate concentrations, then the curve flattens into a plateau at Vmax, because the active sites are saturated and the enzyme concentration is now the limiting factor.
Q10What are Vmax and Km?
Vmax is the maximum rate of reaction, reached when all the active sites are saturated with substrate. Km, the Michaelis-Menten constant, is the substrate concentration at which the rate is half of Vmax; a low Km means a high affinity for the substrate.
Q11How does a competitive inhibitor work, and what happens to Vmax and Km?
It has a shape similar to the substrate and binds to the active site, competing with the substrate. Raising the substrate concentration overcomes it, so Vmax is unchanged but the apparent Km is raised.
Q12How does a non-competitive inhibitor work, and what happens to Vmax and Km?
It binds at a site away from the active site and changes the shape of the enzyme, so the active site is no longer complementary to the substrate. More substrate does not overcome it, so Vmax is reduced and Km is unchanged.
Q13What is an immobilised enzyme, and how are alginate beads made?
An enzyme trapped inside, or attached to, an insoluble support. For calcium alginate, the enzyme is mixed with sodium alginate solution and the mixture is dripped into calcium chloride solution, where the alginate sets into insoluble beads with the enzyme trapped inside.
Q14Give the advantages of using an immobilised enzyme.
It can be separated easily from the product, so the product is not contaminated and no enzyme-removal step is needed; it can be reused many times; the process can be run continuously; and it is more stable to heat and pH, so a higher temperature can be used.
0 of 14 marked as known

Exam-style questions

9 questions, 30 marks. Write your answer, then reveal the mark scheme and tick each point you actually made. Answering honestly here is what makes the rest of the site useful.

StateAO11 mark
State what is meant by the active site of an enzyme.

Mark scheme — tick each point you actually made

Examiner tip: One mark, so give one clear idea. Say where the active site is and what it is complementary to; writing only that it is where the reaction happens is too vague to score.
CompareAO14 marks
Compare the lock-and-key hypothesis with the induced-fit hypothesis for the mode of action of enzymes.

Mark scheme — tick each point you actually made

Examiner tip: A comparison must deal with both hypotheses, so pair your points: the shape of the active site first, then what each hypothesis explains. Naming Fischer and Koshland is a help, but the marks are for the biology.
CalculateAO24 marks
A student investigated the activity of catalase. A fixed mass of potato tissue was added to a fixed volume of hydrogen peroxide solution buffered to pH 7 and kept at 25 °C, and the volume of oxygen collected was recorded every 20 seconds. Time / s: 0, 20, 40, 60, 80, 100. Volume of oxygen / cm3: 0, 4.8, 9.6, 12.6, 14.0, 15.0. (a) Calculate the initial rate of reaction in cm3 per minute, using the readings up to 40 seconds. (b) Calculate the rate of reaction between 60 seconds and 100 seconds, in cm3 per minute. (c) Suggest why the rate of reaction is lower after 40 seconds.

Mark scheme — tick each point you actually made

Examiner tip: A rate is a change divided by the time taken. Take the initial rate from the steepest part of the data, use the two readings either side of the interval in part (b), and give the answer in the unit asked for: 14.4 cm3 per minute, not 0.24.
ExplainAO23 marks
Enzymes from the thermophilic bacterium Thermus aquaticus keep working at temperatures above 70 °C, whereas a human enzyme is denatured above about 45 °C. Explain the difference.

Mark scheme — tick each point you actually made

Examiner tip: Link the structure to the effect: it is the breaking of the bonds that hold the tertiary structure which changes the shape of the active site. Do not write that the human enzyme is killed; a protein is denatured, not killed.
DeduceAO22 marks
In an investigation the Vmax of an enzyme-catalysed reaction was 8.0 arbitrary units per minute. At a substrate concentration of 0.5 mmol/dm3 the rate was 4.0 arbitrary units per minute. Deduce the value of Km for this enzyme and state what that value shows about its affinity for the substrate.

Mark scheme — tick each point you actually made

Examiner tip: A rate of exactly half of Vmax gives the answer straight away, since Km is defined as the substrate concentration at half Vmax. Km is a concentration, so never give it as a rate and never leave off the unit.
ExplainAO26 marks
Explain how a competitive inhibitor and a non-competitive inhibitor each reduce the rate of an enzyme-catalysed reaction, and state the effect of each on Vmax and on Km.

Mark scheme — tick each point you actually made

Examiner tip: Answer in pairs: where the inhibitor binds, whether more substrate overcomes it, then the effect on Vmax and on Km. The mark for Km is the one candidates most often leave out.
DescribeAO34 marks
Describe how a colorimeter can be used to follow the progress of the reaction between amylase and starch, and explain why this is more reliable than judging the end point with iodine solution by eye.

Mark scheme — tick each point you actually made

Examiner tip: Do not stop at writing that a colorimeter is used. The marks are for the correct filter, the blank for calibration, readings at fixed intervals and the use of the initial gradient.
PlanAO34 marks
A student has a solution of the enzyme amylase, a sodium alginate solution and a calcium chloride solution. Plan an investigation to compare the activity of amylase immobilised in alginate with the same concentration of amylase free in solution, using starch as the substrate.

Mark scheme — tick each point you actually made

Examiner tip: A plan is marked for the variables and the method, not for the conclusion. Name the independent variable, at least three controlled variables and exactly how the dependent variable will be measured.
StateAO12 marks
State two advantages of using an immobilised enzyme rather than the same enzyme free in solution in an industrial process.

Mark scheme — tick each point you actually made

Examiner tip: Answer in terms of the process, not the speed of the reaction: immobilisation usually makes an enzyme slower, not faster. Separation, reuse, continuous operation and stability are the marks.

Exam technique for this topic

  • Never write that an enzyme is used up or killed. An enzyme is a protein, not a living thing, and it comes out of the reaction unchanged; high temperature denatures it, and a low temperature only inactivates it.
  • Justify every rate change with the number of enzyme-substrate complexes formed per unit time, and give the cause: more kinetic energy and more successful collisions below the optimum temperature, more active sites available as enzyme concentration rises, or saturated active sites at the plateau of the substrate-concentration curve.
  • In practical questions name the controlled variables and the buffer: the same volume and concentration of enzyme and substrate, the same pH held with a buffer, the same temperature, and readings taken at the same time intervals.
  • With a colorimeter, state the filter and the reason for it (the colour the solution absorbs most strongly, so a red filter for the blue-black starch-iodine complex), calibrate with a blank, and use the gradient of the initial part of the curve. A single absorbance reading is a concentration, not a rate.
  • Learn the two inhibition graphs as a pair. A competitive inhibitor leaves Vmax unchanged and raises the apparent Km; a non-competitive inhibitor lowers Vmax and leaves Km unchanged. A curve that flattens at the same height but starts more gently is competitive inhibition.
  • When comparing Km values, remember that a low Km means a high affinity for the substrate, and state that both enzymes must have been measured at the same temperature and pH because Km is constant only under fixed conditions.

Why students lose marks here

A student wroteA student wrote: 'The enzyme stopped working at 10 °C because it had been denatured by the cold.'
Why it loses marksA low temperature does not break the bonds that hold the tertiary structure, so it cannot denature an enzyme. The reaction is slow because the enzyme and substrate molecules have little kinetic energy, so they collide less often and with less energy and fewer enzyme-substrate complexes form. The effect is completely reversible.
The answer that scoresBelow the optimum temperature the enzyme is inactivated but not denatured: the active site keeps its shape and full activity returns when the mixture is warmed. Only a temperature above the optimum gives enough kinetic energy to break the hydrogen bonds and ionic bonds holding the tertiary structure. That changes the shape of the active site so that it is no longer complementary to the substrate, and the enzyme is permanently denatured.
A student wroteA student wrote: 'I measured the rate of reaction by taking the absorbance reading from the colorimeter after five minutes and recording that value as the rate.'
Why it loses marksA colorimeter measures absorbance, which is proportional to the concentration of the coloured substance in the tube. A rate is a change in amount divided by the time taken, so a single reading can never be a rate, and by five minutes the reaction may already have slowed because substrate has been used up, so the value is far too low.
The answer that scoresTake absorbance readings at fixed time intervals from the moment the enzyme is added, plot absorbance against time, and take the rate of reaction as the gradient of the initial straight part of the curve, where the reaction is fastest. For an iodine end point the rate is 1 / time taken for the starch to disappear. Repeat each experiment and take a mean so that the gradient is reliable.
A laboratory bench set up for an enzyme investigation, with a water bath to control temperature, buffer solutions to control pH, a colorimeter in which the reaction mixture is read at fixed time intervals, and a gas syringe and burettes for measuring the volume of gas produced.
A student wroteA student wrote: 'A competitive inhibitor lowers Vmax, because it blocks the active site so that the substrate cannot get in, and its effect cannot be overcome.'
Why it loses marksBoth parts are wrong. A competitive inhibitor competes with the substrate for the same active site, so raising the substrate concentration overcomes it and the reaction still reaches the original Vmax. What changes is Km, which appears to rise because a higher substrate concentration is now needed to reach half of the maximum rate.
The answer that scoresA competitive inhibitor binds to the active site and raises the apparent Km, but at a high enough substrate concentration the substrate out-competes the inhibitor and the reaction still reaches the original Vmax. A non-competitive inhibitor is the one that lowers Vmax, because it binds away from the active site and changes the shape of the enzyme so that the active site is no longer complementary; adding more substrate does not overcome it, and Km is unchanged.
A student wroteA student wrote: 'Immobilising an enzyme in alginate beads makes it work faster, because the enzyme is held in one place and stays close to the substrate.'
Why it loses marksImmobilisation usually makes an enzyme work more slowly. Trapped enzyme molecules cannot move freely to meet the substrate and the substrate must diffuse into the gel to reach the active sites, so fewer collisions occur, and some active sites are partly blocked by the alginate. The advantages of immobilisation concern separation, reuse and stability, not a faster rate.
The answer that scoresThe same mass of immobilised enzyme normally breaks down less substrate in a given time than the free enzyme, for the reasons above. Immobilisation is still worthwhile because the enzyme can be recovered and reused many times, it does not contaminate the product so no purification step is needed to remove it, it is more stable to heat and to changes in pH so the reaction can be run at a higher temperature, and the process can be run continuously in a packed column.

Where this fits in AS & A Level Biology

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