References only, not question text. Open Official paper ↗ where Cambridge publishes the paper,
or see the official Cambridge past-papers page; for other papers, ask your
school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Enzyme investigations · Core · Paper 4 style · 5 marks
A student in Lahore placed 2 g of grated potato into each of four boiling tubes and added 10 cm³ of hydrogen peroxide solution of different concentrations. Potato contains the enzyme catalase. After 2 minutes she measured the height of the foam. She did three trials at each concentration.Concentration / %: 5 | 10 | 15 | 20Trial 1 foam height / mm: 14 | 27 | 41 | 52Trial 2 foam height / mm: 16 | 29 | 22 | 55Trial 3 foam height / mm: 15 | 31 | 43 | 55
(a)Identify the anomalous result in the table.[1]
(b)Calculate the mean foam height for 15 % hydrogen peroxide, leaving out the anomalous result.[1]
(c)Describe the relationship between the concentration of hydrogen peroxide and the height of the foam.[2]
(d)Name the gas that forms the bubbles in the foam.[1]
Show answer
(a) 22 mm (15 % hydrogen peroxide, trial 2)
B1 22 mm at 15 % – far below the other two trials (41 and 43 mm).
(b) 42.0 mm
(41 + 43) ÷ 2 = 42.0 mm. B1
(c) As the concentration increases, the foam height increases steadily (roughly linear: mean heights 15, 29, 42 and 54 mm, about 12–14 mm more for every 5 % increase).
B1 higher concentration → taller foam; B1 quoting data, e.g. mean 15 mm at 5 % rising to 54 mm at 20 %, or 'steady/roughly linear increase'. Do not accept 'directly proportional' (the line would not pass through the origin).
(d) oxygen
B1 oxygen (catalase breaks hydrogen peroxide into water and oxygen).
Q2Enzyme investigations · Core · Paper 4 style · 7 marks
Two beakers each contained 50 g of chopped apple pulp from apples bought in a Peshawar market. 2 cm³ of pectinase was added to beaker A and 2 cm³ of water to beaker B. Both were kept at 40 °C for 15 minutes and then filtered.Volume of juice from A = 18.5 cm³Volume of juice from B = 7.4 cm³
(a)State why the student set up beaker B with water instead of pectinase.[1]
(b)Calculate the percentage increase in the volume of juice when pectinase was used.[2]
(c)Suggest how pectinase increases the volume of juice released.[2]
(d)State two variables that should be kept constant in this investigation.[2]
Show answer
(a) As a control, to show that the extra juice is caused by the pectinase.
B1 control / comparison / shows effect is due to enzyme.
(b) 150 %
M1 (18.5 − 7.4) ÷ 7.4 × 100; A1 150 %.
(c) It breaks down pectin, which holds plant cell walls together, so cells separate/break open and release more juice.
(d) Any two: mass of apple pulp; volume of liquid added; temperature; time left before filtering; type/ripeness of apple; size of pulp pieces.
1 mark each for any two.
Q3Enzyme investigations · Core · Paper 4 style · 5 marks
Three tubes were kept at 35 °C for 10 minutes and then tested with iodine solution.Tube A: 5 cm³ starch + 1 cm³ amylase → yellow-brownTube B: 5 cm³ starch + 1 cm³ boiled and cooled amylase → blue-blackTube C: 5 cm³ starch + 1 cm³ distilled water → blue-black
(a)Explain the result in tube A.[2]
(b)Explain why tube B stayed blue-black.[2]
(c)State the purpose of tube C.[1]
Show answer
(a) Amylase digested the starch (to maltose/reducing sugar), so no starch was left to turn iodine blue-black.
B1 starch broken down by amylase; B1 so no starch present / iodine stays yellow-brown.
(b) Boiling denatured the amylase, so its active site changed shape and it could not digest starch; starch remained.
B1 boiled enzyme denatured; B1 starch not digested / still present.
(c) A control to show that starch does not break down by itself (without enzyme).
B1 control / starch does not break down without amylase.
Q4Enzyme investigations · Core · Paper 4 style · 4 marks
A student timed how long it took an enzyme reaction to finish at different temperatures. In her table she wrote the column heading 'Time' and wrote values such as '45 s', '62 s' and '80 s' in the cells.
(a)Write a correct column heading for the time column.[1]
(b)State what is wrong with writing '45 s' in each cell of the table.[1]
(c)Name a piece of apparatus to measure exactly 5.0 cm³ of enzyme solution.[1]
(d)Suggest why the student should use a new syringe for the enzyme and for the substrate.[1]
Show answer
(a) Time / s (or time (s))
B1 quantity and unit in the heading.
(b) Units should be in the heading only, not in the cells.
(d) So the enzyme does not mix with the substrate and start reacting before timing begins (no contamination).
B1 avoid contamination / early reaction.
Q5Enzyme investigations · Standard · Paper 4 style · 7 marks
Celery extract was mixed with hydrogen peroxide in a conical flask. The oxygen produced was collected over water in an inverted measuring cylinder.Time / s: 0 | 30 | 60 | 90 | 120 | 150 | 180Volume of oxygen / cm³: 0 | 6.5 | 12.0 | 16.5 | 19.5 | 21.0 | 21.0
(a)Calculate the mean rate of oxygen production during the first 60 s. Give your answer in cm³ per minute.[2]
(b)Calculate the mean rate between 60 s and 120 s, in cm³ per minute.[1]
(c)Explain why no more oxygen was collected after 150 s.[2]
(d)Suggest one source of error in this method and how it could be reduced.[2]
Show answer
(a) 12.0 cm³ per minute
M1 12.0 cm³ in 60 s (1 minute); A1 12.0 cm³/min.
(b) 7.5 cm³ per minute
(19.5 − 12.0) ÷ 1 min = 7.5 cm³/min. B1
(c) All the hydrogen peroxide (substrate) had been broken down, so there was no substrate left for the catalase to act on.
B1 substrate / hydrogen peroxide used up; B1 so no more enzyme–substrate complexes / no more product formed (enzyme still present).
(d) Some oxygen escapes before the bung is put in – use a flask with a side arm/tap funnel so the peroxide is added with the bung already in place (or use a gas syringe).
B1 error (gas lost before bung fitted / bubbles hard to read / reading meniscus of cylinder); B1 matching improvement.
Q6Enzyme investigations · Standard · Paper 4 style · 6 marks
A student mixed 5 cm³ of 1 % starch solution with 5 cm³ of amylase solution. Every 10 s she placed one drop of the mixture into iodine solution on a spotting tile. She recorded the time when the iodine first stayed its original colour.Amylase concentration / %: 0.25 | 0.5 | 1.0 | 2.0Time for starch to disappear / s: 480 | 250 | 130 | 70
(a)Calculate the rate of reaction, 1000 ÷ time, for each amylase concentration. Give each answer to 1 decimal place.[2]
(b)State what colour the iodine solution stays when all the starch has been digested.[1]
(c)Describe the effect of amylase concentration on the rate of reaction.[2]
(d)The drops were tested every 10 s. Explain why the times recorded are not exact.[1]
(d) The starch may have disappeared at any time in the 10 s since the previous test, so each time could be up to 10 s too long.
B1 end point lies between two tests / uncertainty of up to 10 s; test more often.
Q7Enzyme investigations · Standard · Paper 4 style · 6 marks
A protease solution was added to a cloudy milk powder suspension at five temperatures. The time taken for the mixture to become clear was recorded.Temperature / °C: 20 | 30 | 40 | 50 | 60Time to clear / s: 300 | 160 | 85 | 120 | did not clear within 600
(a)Calculate the rate of reaction, 1000 ÷ time, at 40 °C. Give your answer to 1 decimal place.[1]
(b)Explain why the milk powder suspension becomes clear.[2]
(c)Explain why the mixture at 60 °C did not clear within 600 s.[2]
(d)Describe how the student could keep each tube at the stated temperature.[1]
Show answer
(a) 11.8
1000 ÷ 85 = 11.8. B1
(b) Protease breaks down the protein (casein) particles that make it cloudy into soluble amino acids/peptides.
(c) At 60 °C the protease was denatured – the active site changed shape so protein could not bind and be digested.
B1 denatured; B1 active site shape changed / substrate no longer fits.
(d) Stand the tubes in a thermostatically controlled water bath (and check with a thermometer), letting both solutions reach the temperature before mixing.
B1 water bath / thermometer check / pre-incubate solutions.
Q8Enzyme investigations · Standard · Paper 2 style · 7 marks
In Karachi a laundry tested a biological washing powder and a non-biological washing powder on cotton cloths stained with egg and with cooking oil, at 40 °C and at 90 °C.
(a)Name the enzymes in the biological powder that digest (i) an egg stain and (ii) a cooking-oil stain.[2]
(b)Explain why the biological powder removed more of the stains at 40 °C than the non-biological powder.[2]
(c)Explain why the biological powder was no better than the non-biological powder at 90 °C.[2]
(d)Suggest one advantage of washing clothes at 40 °C rather than 90 °C.[1]
Show answer
(a) (i) protease; (ii) lipase
B1 protease; B1 lipase.
(b) The enzymes break the large insoluble stain molecules into small soluble molecules (amino acids; fatty acids and glycerol) which wash out in the water.
(c) At 90 °C the enzymes are denatured, so they can no longer digest the stains.
B1 high temperature denatures enzymes; B1 active site changes shape / no digestion.
(d) Less energy/fuel used to heat the water (cheaper), or delicate fabrics are not damaged.
B1 any one.
Q9Enzyme investigations · Standard · Paper 4 style · 6 marks
Four wells, each 6 mm across, were cut in a dish of starch agar. 0.1 cm³ of a different solution was put in each well and the dish was left for 24 hours, then flooded with iodine solution. The diameter of the clear circle was measured (6 mm means no clear zone beyond the well).0.5 % amylase: 12 mm1 % amylase: 18 mm2 % amylase: 24 mm2 % amylase, boiled: 6 mm
(a)Explain why the agar turned blue-black but there was a clear (yellow-brown) zone around some wells.[2]
(b)Calculate the area of the clear circle for the 1 % amylase. Use area = πr². Give your answer to 1 decimal place.[2]
(c)Explain why the boiled amylase gave no clear zone beyond the well.[1]
(d)Suggest one reason why the diameter is only an estimate of enzyme activity.[1]
Show answer
(a) Iodine turns blue-black with starch in the agar; around the wells amylase diffused out and digested the starch, so there was no starch to stain.
B1 iodine + starch gives blue-black; B1 amylase diffused and digested starch near the well.
(b) 254.5 mm²
M1 r = 18 ÷ 2 = 9 mm; A1 π × 9² = 254.5 mm².
(c) Boiling denatured the amylase so it could not digest starch.
B1 denatured enzyme.
(d) Clear zones may not be perfect circles / edge of the zone is blurred / diffusion rate also affects size.
B1 any one sensible.
Q10Enzyme investigations · Standard · Paper 1 style · 1 mark
(a)Starch solution and amylase were kept at 37 °C until all the starch had been digested. Samples were then tested with iodine solution and with Benedict's solution (heated). Which results are expected?[1]
Show answer
(a) B iodine yellow-brown; Benedict's brick red
No starch remains, so iodine stays yellow-brown; starch has become maltose, a reducing sugar, so Benedict's turns brick red. C would only be seen if some starch were still undigested.
Q11Enzyme investigations · Standard · Paper 4 style · 6 marks
Maize grains were soaked for 2 days so they began to germinate. Some were boiled. All grains were cut in half and placed cut side down on starch agar. After 24 hours the agar was flooded with iodine solution.
(a)Predict what the agar around the germinating seeds looked like after adding iodine, and explain.[3]
(b)Explain the purpose of the boiled seeds.[2]
(c)Suggest why the seeds were cut in half and placed cut side down.[1]
Show answer
(a) A clear yellow-brown area around each seed; germinating seeds make amylase, which diffused into the agar and digested the starch; the rest of the agar was blue-black.
B1 yellow-brown/clear zone round seed; B1 amylase produced by germinating seed; B1 starch digested near seed.
(b) A control: boiling kills the seeds and denatures their enzymes, showing that the clear zones are due to enzymes from living seeds.
B1 control; B1 enzymes denatured / no amylase activity.
(c) So enzymes from inside the seed can diffuse directly into the agar (the seed coat does not block them).
B1 contact of inner tissue with agar.
Q12Enzyme investigations · Challenge · Paper 4 style · 6 marks
Small discs of filter paper soaked in catalase extract sink when dropped into hydrogen peroxide solution. As oxygen is produced, bubbles stick to the disc and it floats to the surface. Hydrogen peroxide solutions of up to 5 % are available, with distilled water, a hole punch, beakers, forceps, a stopwatch and a water bath.
(a)Plan an investigation to find out how the concentration of hydrogen peroxide affects the activity of catalase, using the floating disc method described above.[6]
Show answer
(a) A plan that changes peroxide concentration, keeps other variables constant, measures time for discs to float, repeats, and states how results show activity.
1 mark each, max 6: use at least 5 concentrations (e.g. 0, 1, 2, 3, 4 %), made by diluting with distilled water; same volume of peroxide in each beaker (e.g. 25 cm³); same size discs (cut with same hole punch) soaked in the same catalase extract for the same time; same temperature (water bath, e.g. 25 °C) / same pH (buffer); place disc at bottom with forceps and start stopwatch; record time for disc to rise to the surface; repeat at least 3 times per concentration and calculate a mean; rate = 1 ÷ time / shorter time = higher activity; control with 0 % peroxide or boiled-extract disc; safety: eye protection/gloves with peroxide.
Q13Enzyme investigations · Challenge · Paper 4 style · 7 marks
Lactase breaks down lactose in milk into glucose and galactose. A student added 1 cm³ of lactase to 10 cm³ of milk in five tubes, each with a buffer of different pH, at 30 °C. After 10 minutes she dipped a glucose test strip into each tube and read the glucose concentration from a colour chart.pH: 4 | 5 | 6 | 7 | 8Glucose reading / arbitrary units: 0.5 | 2.0 | 5.5 | 6.5 | 3.0
(a)State the pH at which lactase was most active in this investigation.[1]
(b)Lactose is a reducing sugar. Explain why glucose test strips were used instead of Benedict's solution.[2]
(c)The true optimum pH may not be exactly 7. Describe how the student could find the optimum pH more precisely.[2]
(d)Explain why the glucose reading at pH 4 was very low.[2]
Show answer
(a) pH 7
B1 pH 7 (highest glucose reading, 6.5).
(b) Milk already contains lactose, which gives a positive Benedict's test, so Benedict's cannot show whether lactose has been broken down; the strips detect only glucose, the product.
B1 lactose/milk would also turn Benedict's orange/red (positive before and after); B1 strips specific to glucose / product of the reaction.
(c) Repeat using smaller pH intervals around the peak, e.g. pH 6.0, 6.2, 6.4 … 8.0, with buffers, and repeat each to get means.
B1 narrower range between pH 6 and 8; B1 smaller intervals (e.g. 0.2 or 0.5) / repeats.
(d) At pH 4 the enzyme's shape/active site is changed (denatured), so lactose no longer fits and few enzyme–substrate complexes form.
B1 active site shape changed / denatured; B1 substrate cannot bind / fewer complexes.
Q14Enzyme investigations · Challenge · Paper 4 style · 8 marks
A student investigated catalase in potato. He cut potato pieces by eye, dropped them into 10 cm³ of hydrogen peroxide in a measuring cylinder and measured the foam height with a ruler 'when it looked highest'. The laboratory windows were open and the room temperature changed from 24 °C to 31 °C during the lesson. He did one trial for each condition.
(a)When a classmate repeated one condition twice using the same method, the foam heights were 48 mm and 36 mm. Calculate the percentage by which the second result is lower than the first.[2]
(b)Identify three sources of error in the student's method and suggest an improvement for each.[6]
Show answer
(a) 25 %
M1 (48 − 36) ÷ 48 × 100; A1 25 %.
(b) Pieces cut by eye – cut with a cork borer and ruler to the same size/mass; foam measured while still rising – measure at a fixed time (e.g. exactly 60 s); room temperature changed – use a water bath; only one trial – repeat three times and calculate a mean; foam surface uneven – measure at eye level / use a narrow measuring cylinder.
B1 each error and B1 each matching improvement, max 3 pairs.
Q15Enzyme investigations · Challenge · Paper 4 style · 7 marks
2 g of liver was added to 20 cm³ of hydrogen peroxide of different concentrations. The volume of oxygen collected in a gas syringe in 2 minutes was recorded.Concentration / %: 1 | 2 | 4 | 6 | 8 | 10Volume of oxygen in 2 min / cm³: 4.0 | 8.1 | 15.8 | 20.5 | 21.9 | 22.0
(a)Calculate the rate of oxygen production at 4 % hydrogen peroxide, in cm³ per minute.[1]
(b)Describe the results.[2]
(c)Explain why increasing the concentration above 8 % has almost no effect.[2]
(d)Predict how the results at 10 % would change if 4 g of liver were used instead of 2 g, and explain.[2]
Show answer
(a) 7.9 cm³ per minute
15.8 ÷ 2 = 7.9 cm³/min. B1
(b) Volume increases steeply and almost proportionally from 1 % to 4 %, then levels off above about 6–8 % at about 22 cm³.
B1 increase at low concentrations; B1 levels off/plateau with data (about 22 cm³ from 8 %).
(c) All the active sites of the catalase are occupied (enzyme saturated), so the amount of enzyme now limits the rate.
B1 active sites all in use / saturated; B1 enzyme concentration is limiting factor.
(d) More oxygen (higher plateau), because there are more enzyme molecules / active sites available.
B1 more oxygen / higher rate; B1 more catalase / active sites.
Q16Enzyme investigations · Challenge · Paper 2 style · 6 marks
Equal volumes of hydrogen peroxide were added to samples of different tissues. Oxygen was collected for 5 minutes.Tissue | mass / g | oxygen in 5 min / cm³liver | 2.0 | 46potato | 5.0 | 30apple | 4.0 | 6.0boiled liver | 2.0 | 0
(a)Calculate the rate of oxygen production per gram of tissue per minute for liver, potato and apple.[3]
(b)Explain why the volumes have to be divided by the mass of tissue before the tissues are compared.[1]
(c)Suggest why liver produced the most oxygen per gram.[2]
Show answer
(a) liver 4.6; potato 1.2; apple 0.3 cm³ per g per min
(b) Different masses were used; more tissue contains more catalase, so per gram makes the comparison fair.
B1 masses differ / standardise.
(c) Liver cells contain a lot of catalase because liver cells carry out many reactions (metabolism) that produce hydrogen peroxide, which is toxic and must be removed.
B1 more catalase in liver; B1 to remove toxic hydrogen peroxide made in metabolism.
Q17Enzyme action & specificity · Core · Paper 1 style · 1 mark
(a)Which statement defines an enzyme?[1]
Show answer
(a) A a protein that acts as a biological catalyst
Enzymes are proteins that speed up reactions and are not changed. B is wrong because enzymes are proteins, not carbohydrates.
Q18Enzyme action & specificity · Core · Paper 2 style · 4 marks
(a)State what is meant by the term substrate.[1]
(b)State what is meant by the term active site.[1]
(c)Name the products formed when lipase acts on fat.[2]
Show answer
(a) The substance (molecule) that an enzyme acts on.
B1
(b) The region of the enzyme molecule where the substrate binds.
B1 part of enzyme where substrate fits/binds.
(c) fatty acids and glycerol
B1 fatty acids; B1 glycerol.
Q19Enzyme action & specificity · Standard · Paper 1 style · 1 mark
(a)Amylase digests starch but does not digest protein. What is the reason?[1]
Show answer
(a) B The shape of protein does not fit the active site of amylase.
The active site of an enzyme is complementary in shape to one substrate, so only starch fits amylase. A is wrong because digestion happens outside cells in the gut.
Q20Enzyme action & specificity · Standard · Paper 2 style · 4 marks
In the human gut, protease in the stomach digests the protein in a meal of daal and roti, while amylase from the pancreas digests the starch.
(a)Explain why each enzyme can catalyse only one type of reaction.[4]
Show answer
(a) The active site has a specific shape; only a substrate with a complementary shape fits; it forms an enzyme–substrate complex; the reaction happens and products leave; other molecules do not fit so are not acted on.
1 mark each, max 4: active site specific shape; complementary to substrate; enzyme–substrate complex forms; products released / enzyme unchanged; other substrates do not fit.
Q21Enzyme action & specificity · Standard · Paper 2 style · 5 marks
(a)Complete the information by naming the substrate or product for each enzyme.(i) amylase: starch → ?(ii) protease: ? → amino acids(iii) catalase: hydrogen peroxide → ? + ?[3]
(b)Enzymes are described as 'biological catalysts'. Explain what is meant by a catalyst.[2]
Show answer
(a) (i) maltose (simple reducing sugar); (ii) protein; (iii) water + oxygen
B1 each.
(b) A substance that speeds up a chemical reaction and is not changed by the reaction.
B1 speeds up reaction; B1 not changed / not used up.
Q22Enzyme action & specificity · Challenge · Paper 2 style · 4 marks
(a)In one experiment a single enzyme molecule converted 2.5 × 10⁴ substrate molecules into product each second. Calculate how many substrate molecules it would convert in 5 minutes. Give your answer in standard form.[2]
(b)Explain why one enzyme molecule can convert so many substrate molecules.[2]
Show answer
(a) 7.5 × 10⁶
M1 2.5 × 10⁴ × 300; A1 7.5 × 10⁶.
(b) The enzyme is not used up or changed in the reaction; after products leave the active site it is free to bind another substrate molecule.
B1 not changed/used up (catalyst); B1 active site reused for next substrate.
Q23Effect of temperature · Core · Paper 1 style · 1 mark
(a)An enzyme from the human small intestine was tested at different temperatures. Above 45 °C its activity fell quickly. What is the best explanation?[1]
Show answer
(a) B The enzyme is denatured and its active site changes shape.
High temperature breaks the bonds holding the enzyme's shape, so the substrate no longer fits. A is wrong: molecules move faster, not slower, when heated.
Q24Effect of temperature · Core · Paper 2 style · 5 marks
The activity of an enzyme from human cells and the same type of enzyme from bacteria living in a hot spring were measured.Temperature / °C: 20 | 30 | 40 | 50 | 60 | 70 | 80Human enzyme activity / units: 20 | 45 | 80 | 40 | 5 | 0 | 0Bacterial enzyme activity / units: 5 | 10 | 20 | 35 | 60 | 90 | 40
(a)State the optimum temperature for each enzyme.[2]
(b)Explain why the human enzyme has no activity at 70 °C.[2]
(c)Suggest why the bacterial enzyme is useful in an industrial process carried out at high temperature.[1]
Show answer
(a) human enzyme 40 °C; bacterial enzyme 70 °C
B1 each.
(b) It is denatured; its active site has changed shape so the substrate cannot fit.
B1 denatured; B1 active site shape changed.
(c) It works fastest at high temperature and is not denatured there.
B1
Q25Effect of temperature · Standard · Paper 2 style · 5 marks
(a)Describe and explain how the rate of an enzyme-controlled reaction changes as temperature rises from 10 °C to 70 °C, for an enzyme with an optimum of 40 °C.[5]
Show answer
(a) Rate rises from 10 °C to 40 °C because molecules have more kinetic energy and collide more often, forming more enzyme–substrate complexes; maximum at the optimum (40 °C); above 40 °C rate falls quickly as the enzyme is denatured, active site changes shape, substrate no longer fits; at about 70 °C there is no activity.
1 mark each, max 5: rate increases to 40 °C; more kinetic energy; more collisions / more enzyme–substrate complexes; fastest at optimum; rate falls above optimum; denaturation; active site shape changed / substrate cannot fit; no activity at high temperature / irreversible.
Q26Effect of temperature · Challenge · Paper 4 style · 6 marks
A student investigated the effect of temperature on an enzyme. The enzyme and substrate solutions were placed separately in water baths for 5 minutes, then mixed, and the time for a colour change was recorded.Temperature / °C: 10 | 20 | 30 | 40 | 50 | 60Time / s: 400 | 190 | 95 | 60 | 85 | 300
(a)Calculate the rate, 1000 ÷ time, at each temperature to 1 decimal place.[2]
(b)Calculate how many times faster the reaction was at 30 °C than at 10 °C. Give your answer to 1 decimal place.[1]
(c)State the temperature range in which the optimum lies and describe how to find it more precisely.[2]
(d)Explain why the solutions were left in the water bath for 5 minutes before they were mixed.[1]
(c) Between 30 °C and 50 °C (near 40 °C); repeat at 1–2 °C intervals between 30 °C and 50 °C (e.g. 32, 34 … 48 °C).
B1 30–50 °C; B1 smaller intervals in that range.
(d) So that both the enzyme and the substrate reach the required temperature before the reaction starts.
B1 equilibrate to bath temperature.
Q27Effect of pH · Core · Paper 2 style · 4 marks
The activity of an enzyme was measured at different pH values.pH: 5 | 6 | 7 | 8 | 9 | 10 | 11Activity / units: 4 | 18 | 42 | 60 | 35 | 10 | 0
(a)State the optimum pH for this enzyme.[1]
(b)Describe the effect of pH on the activity of this enzyme.[2]
(c)Suggest which part of the human gut this enzyme might work in. Give a reason.[1]
Show answer
(a) pH 8
B1
(b) Activity rises from pH 5 to a maximum at pH 8 and then falls sharply to zero at pH 11.
B1 increase up to pH 8; B1 decrease above pH 8 with data.
(c) Small intestine (duodenum), because conditions there are slightly alkaline.
B1 small intestine with reason.
Q28Effect of pH · Standard · Paper 1 style · 1 mark
(a)Which enzyme works best at about pH 2?[1]
Show answer
(a) B pepsin in the stomach
The stomach contains hydrochloric acid, and pepsin has an optimum pH of about 2. Trypsin (D) is a protease too but works in the alkaline small intestine.
Q29Effect of pH · Standard · Paper 4 style · 6 marks
Amylase and starch were mixed with buffer solutions of different pH at 35 °C. The time for the starch to disappear (iodine test) was recorded.pH: 4 | 5 | 6 | 7 | 8 | 9Time / s: 360 | 150 | 75 | 60 | 110 | 290
(a)Calculate the rate, 60 ÷ time, at pH 6. Give your answer to 2 decimal places.[1]
(b)State why buffer solutions were used.[1]
(c)Use the results to state the optimum pH and explain how you decided.[2]
(d)State two variables, other than pH, that must be kept the same.[2]
Show answer
(a) 0.80 per minute
60 ÷ 75 = 0.80. B1
(b) To keep the pH of each mixture constant at the chosen value.
B1
(c) pH 7 – it had the shortest time (60 s) / highest rate (1.00 per minute).
B1 pH 7; B1 shortest time / fastest rate.
(d) Any two: temperature; volume/concentration of starch; volume/concentration of amylase; time between iodine tests.
1 mark each.
Q30Effect of pH · Challenge · Paper 2 style · 5 marks
During a meal of chicken karahi with naan, food passes from the mouth to the stomach and then to the duodenum.
(a)Pepsin has an optimum pH of about 2. Explain why pepsin stops digesting protein when food passes into the duodenum.[3]
(b)Salivary amylase digests starch in the mouth. Suggest what happens to its activity when food reaches the stomach, and why.[2]
Show answer
(a) The duodenum is alkaline (bile and pancreatic juice neutralise the acid); the change in pH alters the shape of pepsin's active site / denatures it; protein no longer fits, so no enzyme–substrate complexes form.
B1 duodenum alkaline/neutral (bile, pancreatic juice); B1 active site shape altered / denatured; B1 substrate cannot bind.
(b) It stops working because the stomach is very acidic (about pH 2), far from its optimum near pH 7, so it is denatured.
B1 stops / activity falls; B1 acid pH denatures it.
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
