AQA Enzymes | Maths and Physics Tutor

Amylase is an enzyme that breaks down starch. A student investigated the effect of pH on amylase activity by using a starch agar plate. Six circular wells were cut into the agar plate. Each well contained the same concentration and volume of amylase, and a buffer solution of different pH. The agar plate was then left for 24 hours. The diagram shows the results (a) Describe how the student could have used these results to compare the activity of the enzyme at different pH values. (b) The student concluded that the optimum pH for amylase activity was 7. This conclusion may not be valid. Explain why. (c) Using your knowledge of enzyme structure, explain the result obtained at pH 11. (d) Describe a control that would be necessary for this investigation.

(a) 1. Measure diameter / radius / area of clear zone 2. Detail of method e.g. determine mean diameter of each clear zone / use of graph paper to determine area (b) No measurements at intermediate pH values (c) 1. Enzyme denatured / tertiary structure altered 2. Ionic / hydrogen bonds broken 3. Substrate cannot bind to active site (d) 1. Use of denatured / boiled enzyme 2. At all pH values

Explain why (i) Hydrogen peroxide breaks down at a lower temperature when catalase is present than when it is not present (ii) test tubes A and B became warmer when the reaction was taking place.

(i) (Control) to show that sand did not affect reaction (with ground liver) (ii) 1. Energy in products/water and oxygen less than energy in substrate/ reactants/hydrogen peroxide 2. (Difference) given out as heat / exothermic

The sulphur atoms form bonds between each other, so the cysteine monomers link together in pairs. As a result, the polypeptide chain folds, as shown in the diagram. (i) Explain how this folding helps the lysozyme molecule to work as an enzyme. (ii) Lysozyme breaks down bacterial cell walls but does not affect the walls of plant cells. Explain why.

(i) 1. (Chain folded into) tertiary structure / particular 3-D shape (not just globular) 2. active site formed 3. substrate molecules fit this site 4. reduce activation energy for reaction (ii) 1. Walls made of different materials / peptidoglycan or murein or cellulose 2. specificity of active site / substrate does not fit.

In an investigation, samples of different substances were added to hydrogen peroxide in a series of test tubes. The rate of reaction was measured by recording the rate at which bubbles of oxygen were produced. A scale going from 0 for no bubbles to 5 for the maximum rate of bubbling was used to measure this. The results are shown in the table. PMT Tube Substance added Rate at which bubbles of oxygen were produced Piece of liver 4 Ground liver and sand 5 Sand 0 Piece of cooled, boiled liver 0 Explain the difference between the rate at which bubbles were produced in. (i) tubes A and B; (ii) tubes A and D. (iii) Explain the purpose of tube C.

(i) 1. (Grinding) breaks open cells / increases surface area (of liver) 2. Releases catalase/enzyme/more catalase/ allows more hydrogen peroxide into liver (ii) 1. Heating causes bonds (maintaining tertiary structure) to break 2. Denatures / changes tertiary structure; 3. Active site changed 4. Substrate no longer fits / ES complex not formed (iii) (Control) to show that sand did not affect reaction (with ground liver)

Explain the decrease in the concentrations of potassium ions in the two solutions between 0 and 30 minutes. (i) With inhibitor (ii) Without inhibitor

(i) 1. absorbed by diffusion 2. no energy/ATP available / active transport requires energy/ATP (ii) absorbed by active transport

Explain how (i) raising the temperature to 35 °C affects carbohydrase activity; (ii) decreasing the pH affects carbohydrase activity.

(i) 1. increase in temperature increases kinetic energy 2. increases collisions (between enzyme/active site and substrate) / increases formation of enzyme/substrate complexes 3. increases rate of breakdown of starch /rate of reaction/carbohydrase activity (ii) 4. (decrease in pH) increases H+ ions/protons 5. attach/attracted to amino acids; 6. hydrogen/ionic bonds disrupted/broken; 7. denatures enzyme / changes tertiary structure 8. changes shape/charge of active site; 9. active site/enzyme unable to combine/fit with starch/enzyme-substrate complex no longer able to form 10. decreases rate of breakdown of starch/rate of reaction /carbohydrase activity

In mammals, amino acids are broken down and urea is formed. Urea from animal waste is often used as a natural fertiliser. Soil bacteria secrete an enzyme called urease that breaks down urea into ammonia and carbon dioxide. Some of this ammonia is released into the atmosphere. urea + water carbon dioxide + ammonia Scientists have studied this reaction because it results in the loss of fertiliser. They have produced a substance called NBPT which is added to urea fertiliser. NBPT is an enzyme inhibitor which affects the action of the urease produced by soil bacteria. The graph shows the results of an experiment in which a standard amount of urease and of the enzyme inhibitor NBPT were added to different concentrations of urea solution. Describe and explain the effect on the rate of the reaction of increasing the urea concentration (i) without NBPT present (ii) with NBPT present.

(i) 1. increases up to 20 - 29 units of urea / rate 20-21 2. since urea concentration limiting rate / more urea - enzyme collisions ONCE 3. then (high) constant / levels off 4. since active sites all (continually) occupied 5. (saturated neutral) other named factor limiting e.g. enzyme concentration (ii) 1. increases up to 45-50 units / rate 17-19 since urea concentration limiting rate / more urea - enzyme collisions ONCE; 2. NBPT reduces rate of reaction 3. reduction greater at low concentration of urea than at high concentration 4. competitive inhibitor / competes for active site 5. since complementary shape / similar shape to substrate (NOT same shape) 6. at high concentrations urea competes more successfully for active site / more enzyme - urea collisions

A student carried out an investigation into the mass of product formed in an enzyme-controlled reaction at three different temperatures. Only the temperature was different for each experiment. The results are shown in the graph. Use your knowledge of enzymes to explain (i) why the initial rate of reaction was highest at 55 °C; (ii) the shape of the curve for 55 °C after 20 minutes. (iii) Explain why the curves for 27 °C and 37 °C level out at the same value.

(i) 1. substances/molecules have more (kinetic) energy/moving faster 2. increased collisions 3. enzyme substrate complexes formed (ii) 1. causes denaturation/tertiary structure/shape change 2. H+/ionic bonds break 3. (shape) of active site changed 4. substrate no longer binds/not complementary to (active site) (iii) 1. all substrate changed into product / reaction is complete 2. same amount of product formed 3. same initial substrate concentration

The effect of temperature on the rate of reaction of an enzyme was investigated. A test tube containing the enzyme and a test tube containing the substrate were incubated separately at each of the temperatures being investigated. After 5 minutes, they were mixed and the rate of reaction was determined. The experiment was repeated but, this time, the enzyme and the substrate were left for 60 minutes before they were mixed. The results of the investigation are shown in the graph. (i) The enzyme solution used in this investigation was made by dissolving a known mass of enzyme in a buffer solution. Explain why a buffer solution was used. (ii) Use the graph to describe how incubation time affects the rate of the reaction (iii) The maximum rate of reaction with an incubation time of 60 minutes is less than the maximum rate of reaction with an incubation time of 5 minutes. Explain why

(i) To keep pH the same / optimum pH / so change in pH does not affect reaction (ii) 1. For temperature up to 40 - 50°C has no effect 2. Over temperature (of 40 - 50°C) reduces rate of reaction (iii) 1. Bonds (holding tertiary structure) broken 2. More enzyme denatured / tertiary structure destroyed 3. Active sites lose shape/no longer fit 4. Fewer enzyme-substrate complexes formed

The graph shows the total amount of product formed during an enzyme-controlled reaction. (i) Use your knowledge of how an enzyme works to explain why the rate of reaction decreased with time. (ii) Explain why the initial rate of reaction was measured at each substrate concentration.

(i) 1. Substrate becomes limiting/ falls/ gets less 2. Fewer collisions/ complexes formed (ii) 1. Enables a comparison to be made 2. As the rate/concentration changes as reaction progresses 3. Cells/ catalase may become damaged/affected by heat

This chemical test was carried out on samples A and B. All experimental variables were the same in the testing of the two samples. Both tubes were left for ten minutes to allow the precipitate to settle. The diagram shows the result. Is galactose a reducing sugar? Explain how the results in the diagram support your answer.

(yes) more precipitate in sample B both sugars are reducing sugars/ give a positive test

Use your knowledge of protein structure to explain why enzymes are specific and may be affected by non-competitive inhibitors.

1 each enzyme/protein has specific primary structure / amino acid sequence 2 folds in a particular way/ has particular tertiary structure 3 active site with unique structure 4 shape of active site complementary to/ will only fit that of substrate; maximum of three marks for inhibition, points 5 - 8 5 inhibitor fits at site on the enzyme other than active site 6 determined by shape 7 distorts active site 8 so substrate will no longer fit / form enzyme-substrate complex

Explain how adding excess substrate could overcome the effect of a competitive inhibitor.

1. (More substrate than inhibitor so) more likely to form enzyme-substrate complex 2. more likely for substrate to enter the active site

It is better to use a biosensor than the Benedict's test to measure the concentration of glucose in a sample of blood. Suggest two reasons why.

1. (Only detects glucose whereas) Benedict's detects (all) reducing sugars/named examples 2. Provides a reading / is quantitative / Benedict's only provides a colour / doesn't measure concentration / is qualitative/semiquantitative 3. Is more sensitive / detects low concentration 4. Red colour/colour of blood masks result 5. Can monitor blood glucose concentration continuously

Bacteria produce enzymes which cause food to decay. Explain how vinegar, which is acidic, can prevent the action of bacterial enzymes in some preserved foods.

1. (bacterial) active site/enzymes/proteins denatured / tertiary 3D structure disrupted/changed 2.(ionic) bonds broken 3. no enzyme substrate complex formed / substrate no longer fits

Explain why the rate of reaction of enzyme B is low at pH 5.

1. (change in pH) leads to breaking of bonds holding tertiary structure / changes charge on amino acids 2. enzyme/protein/active site loses shape/denatured 3. substrate will not bind with/fit active site; fewer/no ES complexes formed

Explain how amylase makes it possible for starch to be digested at body temperature.

1. Activation energy reduced 2. starch attached to active site / formation of enzyme substrate complex 3. less energy required to bring (substrate) molecules together / to break bonds 4. reaction occurs in small(er) steps 5. change in shape of enzyme molecule (induced fit) brings molecules together / allows bonds to break / causes overlapping of electron orbits of substrates.

Explain what happens to an enzyme molecule when it is denatured by high temperature.

1. Correctly named bonds broken / water removed 2. tertiary / globular shape of enzyme changed 3. shape of active site affected

Use your knowledge of enzymes to explain how cyanide stops the activity of cytochrome oxidase.

1. Cyanide binds to enzyme molecule away from active site 2. shape of active site changed. 3. cyanide attaches permanently to active site 4. active site blocked.

Use your knowledge of enzymes to explain why the activity of the amalyse decreased.

1. Enzyme (molecules) denatured at 60°C / high temperature, or description of denaturing (e.g. vibration disrupts enzymes) 2. change (in shape) of active site 3. change in tertiary/'3D'structure / 4. hydrogen bonds broken 5. substrates no longer fit 6. loss of activity of enzyme in water bath due to slow denaturing

It has been suggested that it might be possible to use cyanide to kill cancer cells. Cassava plants make a substance called linamarin and also an enzyme that breaks down linamarin to release cyanide. The idea is to use an antibody that attaches only to the cell surface membrane of cancer cells. The enzyme from the cassava plants would be combined with this antibody. Linamarin would then be injected into the person with cancer. Explain how this method would kill the cancer cells, but not other cells.

1. Enzyme + antibody attaches (to membrane) of cancer cells only 2. Enzyme breaks down (injected) linamarin 3. Cyanide released disrupts respiration/metabolism of cancer cells.

A glucose biosensor detects only glucose. Use your knowledge of the way in which enzymes work to explain why.

1. Enzyme/active site has a (specific) tertiary structure 2. Only glucose has correct shape / is complementary / will bind/fit to active site 3. (Forming) enzyme-substrate complex

Use your knowledge of the tertiary structure of enzymes to explain how a non competitive inhibitor could reduce the rate of an enzyme controlled reaction.

1. Inhibitor is a different shape to substrate 2. Binds at position other than active site/allosteric site 3. Alters shape of active site 4. Substrate cannot bind/enzyme-substrate complex not formed

Sucrase does not hydrolyse lactose. Use your knowledge of the way in which enzymes work to explain why.

1. Lactose has a different shape/structure 2. Does not fit/bind to active site of enzyme/sucrase

Molecules A and B inhibit the enzyme in different ways. Explain how each molecule inhibits the enzyme.

1. Molecule A binds at site away from active site / allosteric site 2. Causes enzyme / active site to change shape 1. Molecule B can enter / competes for active site 2. Prevents substrate from entering / no enzyme-substrate complex formed / active site blocked

A drug company produced a new type of insulin. Scientists from the company carried out a trial in which they gave this new type of insulin to rats. They reported that the results of this trial on rats were positive. A newspaper stated that diabetics would benefit from this new drug. Suggest two reasons why this statement should be viewed with caution.

1. Study not carried out on humans / only carried out on rats; Long-term/side effects not known 2. Scientists have vested interest 3. Study should be repeated / further studies / sample size not known

Explain why increasing substrate concentration above the value shown as X fails to increase the rate of the reaction further.

1. Substrate concentration not limiting / enzyme concentration limiting 2. all active sites of enzyme full / enzyme at maximum turnover rate

Explain how substrate C is broken down by the enzyme.

1. Substrate enters active site 2. Complimentary shapes / Lock and Key 3. (Binding) to form enzyme-substrate complex 4. Lowering of activation energy 5. Conformational / shape change 6. Breaking of bonds in substrate 7. Products no longer fit active site and so are released

Gout is a painful condition caused by uric acid crystals in the joints. It is often treated with a drug that inhibits xanthine oxidase. The diagram shows a molecule of xanthine and a molecule of this drug.

1. Xanthine similar shape to drug 2. Drug fits active site/competes for active site/is a competitive inhibitor 3. Less/no uric acid formed

Explain the effect of the non-competitive inhibitor.

1. inhibitor attaches to enzyme away from the active site 2. changes shape of active site 3. prevents formation of enzyme-substrate complex

Many reactions take place in living cells at temperatures far lower than those required for the same reactions in a laboratory. Explain how enzymes enable this to happen.

1. lowers activation energy 2. relevant mechanism e. g. brings molecules close together / reaction in smaller steps / change in charge distribution / proton donation or acceptance / induced fit ensuring substrates brought in correct sequence 3. including relevant reference to active site;

Explain why an increase in concentration of phenol solution from 2.0 to 2.5 mmol dm-3 has no effect on the rate of the reaction without inhibitor.

1. maximum rate at which enzyme can combine with substrate / form enzyme-substrate complexes / substrate no longer limiting / 2. enzyme now the limiting factor 3. All (active sites of) enzyme saturated with substrate

A decrease in temperature decreases the kinetic energy of molecules in a solution. Explain how a decrease in temperature decreases the rate of an enzyme-controlled reaction.

1. molecules moving less/slower 2. reduces chance of collision (between enzyme and substrate)/of enzyme-substrate complexes being formed;

Enzyme A is present in some washing powders used for cleaning clothes. Use the graph to suggest why enzyme A would be of more use in washing clothes than enzyme B.

1. more resistant to changes in pH and washing conditions variable/ works in alkaline pH and washing powders alkaline

Explain how the shape of an enzyme molecule is related to its function.

1. specific 3D tertiary structure/shape 2. substrate complementary shape 3. substrate (can bind) to active site/ can fit into each active site

Urea breaks hydrogen bonds. Explain how the addition of urea would affect the rate of an enzyme-controlled reaction.

1. these bonds hold/maintain tertiary/globular structure (of enzyme) enzyme denatured/tertiary structures destroyed 2. (shape of) active site distorted/changes substrate no longer fits/enzyme-substrate complex not formed

Diabetes mellitus is a disease that can lead to an increase in blood glucose concentration. Some diabetics need insulin injections. Insulin is a protein so it cannot be taken orally. Suggest why insulin cannot be taken orally.

Broken down by enzymes / digested / denatured (by pH) too large to be absorbed

Explain how inhibitors affect the rate of enzyme-controlled reactions.

Competitive 2 Similarity of shape of inhibitor and substrate 3 Inhibitor can enter/bind with active site (of enzyme) Non-competitive 4 Affect/bind to enzyme other than at active site 5 Distorts shape of active site Inhibitors 6 Prevent entry of/binding of substrate to active site 7 Therefore fewer/no enzyme-substrate complexes formed

Alcohol dehydrogenase is an enzyme found in the liver. It normally breaks down ethanol (C2H5OH) into less harmful products. About 50 deaths each year occur following ingestion of a compound called ethylene glycol (C2H4(OH)2) which is found in antifreeze. This compound is not lethal, but it is broken down by alcohol dehydrogenase into highly toxic oxalic acid. Giving a large dose of ethanol as quickly as possible can treat ethylene glycol poisoning. Use your knowledge of enzyme activity to suggest how this treatment may counteract the effects of ethylene glycol poisoning.

Competitive inhibition 1. Ethanol/ethylene glycol compete for same active site 2. Molecules similar shape (not same)/both complementary to/both fit active site 3. Prevents/slows production or build up of oxalic acid/toxic products 4. Ethylene glycol excreted (without causing death)

Describe and explain how an increase in temperature affects the rate of an enzyme controlled reaction.

Temperature 1. Rate of reaction increases 2. Increasing temperature increases rate of movement of molecules/ kinetic energy 3. Collide more often/substrate enters active site more often/ 4. more enzyme-substrate complexes formed; Up to optimum 1. Rate of reaction decreases 2. High temperatures cause denaturation/loss of tertiary structure/3D structure 3. By breaking specified bonds (not peptide bond) 4. Active site altered/substrate cannot bind/fit