IGCSE Biology Paper 4 (Theory / Extended)

Topic 5: Enzymes -- Challenge Exam 3

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0610

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

This paper covers the whole of Topic 5. Like a real Cambridge paper, the seven questions range across every sub-topic — catalysts and the active site, temperature and pH, and investigating enzyme activity — and they are deliberately mixed rather than grouped. All three Topic 5 papers do; they differ in the angle they come at it from, not in what they cover.

Question 1 — Planning It From Nothing

Total: 12 marks

You are given: amylase solution, 1% starch solution, iodine solution, white spotting tiles, syringes, test tubes, water baths set to a range of temperatures, buffer solutions, thermometers and a stopclock. You are asked to investigate how temperature affects the activity of amylase.

(a) [6]

Describe a method you would use. Your answer should make clear how you would measure the activity and how you would ensure the investigation is reliable.

Model Answer — 1(a)

use water baths at a range of at least five temperatures, for example 10, 20, 30, 40 and 50 °C [1]

measure equal volumes of starch and amylase with syringes, e.g. 5 cm³ of each, and add the same volume of pH 7 buffer to every tube [1]

place the starch and the amylase in the water bath separately for five minutes to equilibrate before mixing them [1]

mix, start the stopclock, and every 30 s remove a drop with a clean syringe and add it to a drop of iodine on the spotting tile [1]

the end-point is when the iodine stays orange-brown, showing all the starch has been digested; record the time [1]

repeat three times at each temperature, take a mean time, and calculate rate as 1000 ÷ mean time [1]

⚠ If you missed marks here: Method answers are marked against a checklist, so write in the checklist order and nothing gets forgotten. The two marks most often dropped are equilibrating before mixing and buffering the pH. “Repeat to make it reliable” only scores if you also say a mean is taken, and the final mark needs the conversion to rate — a graph of raw times comes out upside down.

1 mark — range of at least five temperatures with values given 1 mark — equal measured volumes of starch, amylase and buffer 1 mark — solutions equilibrated separately in the water bath before mixing 1 mark — samples tested with iodine at regular intervals 1 mark — end-point stated as iodine remaining orange-brown, and time recorded 1 mark — repeats with a mean, and conversion to rate

(b) [3]

State three variables that must be controlled, and for each explain why.

Model Answer — 1(b)

the volume and concentration of starch, e.g. 5 cm³ of 1%, because more substrate would take longer to digest regardless of temperature [1]

the volume and concentration of amylase, because more enzyme means more active sites and a faster reaction [1]

the pH, held constant with a buffer, because pH alters the shape and fit of the active site and would confound the effect of temperature [1]

⚠ If you missed marks here: Each mark needs a named variable with a reason, and quantities make the answer stronger at almost no cost. Do not offer “temperature” here — it is the independent variable. “Keep everything else the same” earns nothing.

1 mark — volume and concentration of starch, with reason 1 mark — volume and concentration of amylase, with reason 1 mark — pH held constant with a buffer, with reason

Predict the shape of the graph of rate against temperature that you would obtain, and explain the biological reason for each part of it.

Model Answer — 1(c)

the rate would rise to a peak at the optimum and then fall steeply to zero [1]

below the optimum, molecules gain kinetic energy, so effective collisions between starch and the active site are more frequent and more enzyme–substrate complexes form [1]

above the optimum the amylase is denatured: the shape of the active site changes so starch is no longer complementary to it, no complexes form and the rate falls to zero [1]

⚠ If you missed marks here: Predicting a curve is easy; explaining both halves separately is where the marks are. A single blended sentence such as “the rate rises until it gets too hot” describes the graph without explaining it and is worth at most one mark.

1 mark — prediction of a rise to a peak followed by a steep fall to zero 1 mark — rising side explained by kinetic energy, effective collisions and complexes 1 mark — falling side explained by denaturation and the loss of complementary fit

Question 2 — Four Faults and a Missing Tube

Total: 12 marks

A student submits this plan. “To find the effect of pH on catalase, I will put a piece of potato into a boiling tube of hydrogen peroxide, add a few drops of acid or alkali to change the pH, and count the bubbles for one minute. I will do each pH once. I will use whichever potato pieces are left in the tray, and work at room temperature. The pH that gives the most bubbles is the optimum.”

(a) [8]

Identify four faults in this plan. For each, explain the effect it would have on the results and state how it should be corrected.

Model Answer — 2(a)

Fault 1: adding a few drops of acid or alkali gives an unknown pH that drifts during the reaction [1]; use buffer solutions of known pH, the same volume in every tube [1]

Fault 2: counting bubbles is unreliable because bubbles vary in size, so equal counts do not mean equal volumes [1]; collect the oxygen in a gas syringe or inverted measuring cylinder and record the volume [1]

Fault 3: potato pieces “left in the tray” differ in size and may come from different potatoes, so the mass, surface area and catalase concentration vary [1]; cut discs of equal size with a cork borer and ruler from a single potato [1]

Fault 4: a single trial at each pH cannot reveal an anomaly or show reliability [1]; repeat at least three times at each pH and take a mean (accept: room temperature is uncontrolled — use a water bath) [1]

⚠ If you missed marks here: Eight marks split into four pairs, and each pair needs the fault and its correction. The commonest way to lose half of them is to list four faults quickly and never say what should be done instead. Be specific: “be more careful” and “avoid human error” are never credited.

1 mark — fault 1 identified: unknown, drifting pH 1 mark — correction 1: buffers of known pH, equal volumes 1 mark — fault 2 identified: bubble counting is unreliable 1 mark — correction 2: measure the volume of gas collected 1 mark — fault 3 identified: potato pieces vary in size and source 1 mark — correction 3: equal discs cut from one potato with a cork borer 1 mark — fault 4 identified: only one trial, so no anomalies or reliability 1 mark — correction 4: three repeats and a mean (or a water bath for temperature)

(b) [2]

Describe the control experiment the student has omitted, and state the result you would expect from it.

Model Answer — 2(b)

a tube containing boiled potato discs with hydrogen peroxide and buffer, otherwise identical [1]

expected result: little or no oxygen released, because the catalase has been denatured — showing that the oxygen in the other tubes was produced by the enzyme [1]

⚠ If you missed marks here: A control experiment is a whole extra tube, not a factor held constant. The second mark is for saying what the control proves: without it, a critic could argue the oxygen came from the hydrogen peroxide decomposing on its own.

1 mark — boiled potato control tube described 1 mark — expected result of little or no oxygen, and what it shows

The student writes: “The pH that gives the most bubbles is the optimum.” Explain one reason why this statement is unsafe even if the method is corrected.

Model Answer — 2(c)

the optimum can only be located as precisely as the intervals tested — if pH values are two units apart, the true optimum may lie between two of them [1]

so the correct statement is that the optimum lies between the neighbouring values tested, and narrower intervals around the peak would be needed to locate it [1]

⚠ If you missed marks here: Notice that this fault survives every other correction, which is why it is asked separately. A conclusion can never be more precise than the intervals it was measured at, and saying so explicitly is a reliable source of marks in planning questions.

1 mark — recognition that precision is limited by the interval between pH values tested 1 mark — correct statement of the conclusion, with narrower intervals suggested

Question 3 — Is Denaturation Really Permanent?

Total: 12 marks

A student reads that denaturation is irreversible and decides to test it. She takes six identical samples of a protease with an optimum of 40 °C. Each is held at a stated temperature for 15 minutes, then all six are placed in a water bath at 40 °C for 10 minutes before being assayed with the same protein substrate under identical conditions.

sample	holding temperature / °C	activity when assayed at 40 °C / units
A	4	96
B	25	98
C	40	100
D	55	74
E	70	6
F	90	0

(a) [3]

Explain why sample C is an important part of this design, and why all six samples were assayed at 40 °C.

Model Answer — 3(a)

sample C is held at the optimum, so it acts as the control against which the others are compared, and it defines 100% activity [1]

assaying all six under identical conditions means the assay temperature cannot itself affect the result [1]

so the only variable is the temperature at which each sample was held, and any difference must be caused by lasting change to the enzyme [1]

⚠ If you missed marks here: The design is the answer here. If each sample had been assayed at its holding temperature, a low value could mean either damage or simply low kinetic energy, and the experiment could not distinguish them — which is precisely what it was built to do.

1 mark — sample C identified as the control held at the optimum, defining 100% 1 mark — identical assay conditions mean the test temperature cannot affect the result 1 mark — the only variable is the holding temperature, so differences show lasting change

(b) [4]

State what the results show about the reversibility of denaturation, using figures, and explain the biological reason.

Model Answer — 3(b)

samples held at 4 and 25 °C recover essentially full activity (96 and 98 units), so cooling causes no lasting damage [1]

samples held at 70 and 90 °C give 6 and 0 units even after 10 minutes at the optimum, so the loss is not reversed by returning to favourable conditions [1]

high temperature breaks the bonds holding the folded protein chain, and the chain does not re-fold correctly [1]

so the active site does not regain its shape, the substrate is no longer complementary to it, and no enzyme–substrate complexes can form [1]

⚠ If you missed marks here: Both halves are needed: the cold samples prove the method can detect recovery, and the hot samples prove none occurs. Answers that quote only the hot samples miss the logic — without the cold controls, a critic could claim the assay itself was faulty.

1 mark — cold samples recover full activity, with figures — cold causes no lasting damage 1 mark — hot samples do not recover, with figures — the loss is not reversed 1 mark — heat breaks the bonds holding the fold and the chain does not re-fold 1 mark — active site does not regain its shape, so no complementary fit and no complexes

Sample D retains 74 units of activity. Suggest what is happening in this sample and how the student could investigate it further.

Model Answer — 3(c)

55 °C is above the optimum, so some of the enzyme molecules have been denatured while others remain intact [1]

the measured activity reflects the proportion of molecules that survived, not a partial slowing of every molecule [1]

she could hold samples at 55 °C for different lengths of time — 5, 10, 20, 40 minutes — and assay each, to show that the proportion denatured increases with exposure time [1]

⚠ If you missed marks here: The idea that a population of molecules is denatured gradually, rather than each molecule working at half speed, is a genuinely challenging one and it is what the 74 units are there to prompt. The suggested investigation must change one variable only: the time of exposure.

1 mark — recognition that some molecules are denatured while others are intact at 55 °C 1 mark — activity reflects the proportion of surviving molecules, not partial slowing of each 1 mark — further investigation: vary the exposure time at 55 °C and assay each sample

(d) [2]

A classmate says the results prove that the optimum temperature of this enzyme is 40 °C. Explain why this experiment cannot show that.

Model Answer — 3(d)

every sample was assayed at 40 °C, so the experiment measures how much enzyme survived each holding temperature, not how fast the enzyme works at different temperatures [1]

to find the optimum the rate would have to be measured at a range of temperatures, not at one fixed temperature [1]

⚠ If you missed marks here: The 40 °C figure comes from the stem, not from the data, and the design cannot test it. Asking “what did this experiment actually vary?” before drawing any conclusion is a habit worth building.

1 mark — recognition that the experiment measures survival, not activity at each temperature 1 mark — correct design stated: measure the rate at a range of temperatures

Question 4 — Enzymes on a Bead

Total: 12 marks

Some people produce very little lactase and cannot digest lactose, the sugar in milk. Lactose-free milk is made industrially by attaching lactase to small beads packed into a column, and passing milk slowly down the column. The beads are used continuously for several weeks before being replaced. The lactase used comes from a fungus and has an optimum of 45 °C and pH 6.5.

(a) [3]

Name the products of the reaction, and explain in terms of the active site why this lactase has no effect on the protein or fat in the milk.

Model Answer — 4(a)

the products are glucose and galactose [1]

the active site of lactase has a specific shape, and only lactose is complementary to it [1]

protein and fat molecules cannot bind, so no enzyme–substrate complex forms and they are not broken down — this is specificity [1]

⚠ If you missed marks here: The names of the two products are given in most textbooks and are worth learning, but if you cannot recall them the explanation marks are still fully available. The chain must run through the active site: “lactase only works on lactose” restates the question rather than answering it.

1 mark — glucose and galactose named as products 1 mark — active site has a specific shape complementary only to lactose 1 mark — other molecules cannot bind, so no complex forms — specificity

(b) [3]

Explain why the same beads can be used for several weeks.

Model Answer — 4(b)

the enzyme is a catalyst and is not changed by the reaction [1]

each lactase molecule binds a lactose molecule, forms products, releases them and is immediately free to bind another [1]

so a single molecule catalyses the reaction many thousands of times and is not consumed as the milk passes through [1]

⚠ If you missed marks here: This is the definition of a catalyst turned into an industry. Answers that suggest the enzyme is slowly replaced from the milk, or that the beads release more enzyme over time, have missed the clause “not changed by the reaction” entirely.

1 mark — enzyme is a catalyst, not changed by the reaction 1 mark — cycle described: binds, forms products, releases, binds again 1 mark — each molecule works many times and is not consumed

The milk is passed through the column at 8 °C rather than at 45 °C. Explain the disadvantage of this, and suggest why the company does it anyway.

Model Answer — 4(c)

at 8 °C the molecules have far less kinetic energy, so effective collisions between lactose and the active site are much less frequent and the reaction is slow [1]

the milk must therefore pass through the column much more slowly, or a longer column is needed [1]

the company does it because milk kept at 45 °C would spoil rapidly as bacteria multiply, so keeping it cold protects the product [1]

⚠ If you missed marks here: A suggest mark rewards a sensible reason outside the immediate biology. The structure to copy is: state the cost, quantify it in terms of the process, then give the compensating benefit.

1 mark — at 8 °C less kinetic energy, so fewer effective collisions and a slow reaction 1 mark — consequence for the process: slower flow rate or a longer column needed 1 mark — reason: warm milk would spoil because bacteria multiply rapidly

(d) [3]

The beads are washed with a hot cleaning solution at pH 11 between production runs. Explain why this would eventually make them useless, and suggest one change that would prolong their life.

Model Answer — 4(d)

high temperature and pH 11 both denature the lactase [1]

the shape of the active site changes so lactose is no longer complementary to it, and the change is permanent, so activity is lost with each wash [1]

suggestion: wash at a lower temperature and a pH closer to the enzyme optimum, or use a shorter washing time (accept: use an enzyme from an organism tolerant of those conditions) [1]

⚠ If you missed marks here: This part asks you to apply denaturation to a commercial problem rather than a test tube, and the reasoning is identical. Note that both factors in the wash — heat and alkali — act on the same target: the shape of the active site.

1 mark — heat and pH 11 both denature the lactase 1 mark — active site changes shape permanently, so activity is lost with each wash 1 mark — a sensible change suggested to prolong the life of the beads

Question 5 — From Amino Acids to Active Sites

Total: 10 marks

Enzymes are proteins. A protein is built from amino acids joined into a chain, and the order of those amino acids is determined by the sequence of bases in DNA. Two people produce versions of the same digestive enzyme that differ at a single position in the chain. Person 1 digests the substrate normally; person 2 cannot digest it at all, although the enzyme is produced in normal amounts.

(a) [3]

Name the small molecules from which an enzyme is built, name the chemical elements always present in an enzyme, and state one chemical test that would confirm that an unknown sample is a protein.

Model Answer — 5(a)

built from amino acids [1]

elements always present: carbon, hydrogen, oxygen and nitrogen (sulfur is present in some proteins only) [1]

add biuret solution; a positive result is a colour change from blue to purple [1]

⚠ If you missed marks here: Nitrogen is the element that identifies a protein, since no carbohydrate or fat contains it — but sulfur must be described as present in some proteins, not all. The biuret test requires no heating, unlike the Benedict’s test.

1 mark — amino acids named 1 mark — C, H, O and N named, with sulfur correctly qualified if mentioned 1 mark — biuret test named with the blue to purple colour change

(b) [4]

Explain fully why person 2 cannot digest the substrate.

Model Answer — 5(b)

the order of the amino acids determines how the protein chain folds [1]

the folding produces the active site, so a change in the chain can alter the shape of the active site [1]

the substrate is then no longer complementary to the active site and cannot bind [1]

no enzyme–substrate complex forms, so the reaction is not catalysed and the substrate is not digested [1]

⚠ If you missed marks here: Four marks, four links, in order. The trap is the word denatured: this enzyme was assembled with the wrong shape and was never damaged, whereas denaturation is the loss of shape of an enzyme that had folded correctly.

1 mark — amino acid order determines the folding of the chain 1 mark — folding determines the shape of the active site 1 mark — substrate no longer complementary and cannot bind 1 mark — no enzyme–substrate complex forms, so no reaction

Person 2 avoids the problem by taking a tablet containing the missing enzyme with meals. Explain why the tablet has a coating that dissolves only in the small intestine and not in the stomach.

Model Answer — 5(c)

the stomach has a pH of about 2, which is far from the optimum of this enzyme [1]

the enzyme would be denatured there: the shape of its active site would change so the substrate could no longer bind [1]

the coating protects it until it reaches the small intestine, where the pH of about 8 is close to its optimum and it can work [1]

⚠ If you missed marks here: This is applied pH, and the answer must run through denaturation rather than stopping at “the acid would destroy it”. Notice that the damage would be permanent, which is why protecting the enzyme is worth the cost of a coating.

1 mark — stomach pH of about 2 is far from the enzyme optimum 1 mark — enzyme would be denatured, with the active site changing shape 1 mark — coating protects it until the small intestine, where the pH suits it

Question 6 — Comparing Two Enzymes Fairly

Total: 12 marks

A company sells two protease preparations, X and Y, for use in a food process carried out at 30 °C and pH 7. A technician is asked to decide which preparation is more suitable. She has both preparations, a protein substrate, water baths, buffers, syringes and the apparatus for a milk-clearing test in which the time for a cloudy protein suspension to become clear is recorded.

(a) [4]

Describe how she should compare the two preparations fairly. State the independent variable, the dependent variable and two control variables.

Model Answer — 6(a)

independent variable: the preparation used, X or Y [1]

dependent variable: the time for the suspension to become clear, converted to rate as 1 ÷ time [1]

control variable 1: the same volume and concentration of the protein substrate in every tube [1]

control variable 2: the same temperature (30 °C in a water bath) and the same pH (pH 7 buffer) for both preparations [1]

⚠ If you missed marks here: Naming the three types of variable correctly is worth marks on its own, and mislabelling them undermines everything after. The dependent variable should be stated as a measurement, not as “the result”, and converting time to rate is what makes the comparison readable.

1 mark — independent variable correctly identified as the preparation used 1 mark — dependent variable identified as the time to clear, with conversion to rate 1 mark — substrate volume and concentration named as controlled 1 mark — temperature and pH named as controlled, with how they are held

(b) [3]

Explain why she must also check that the two preparations contain the same concentration of enzyme, and describe what would happen to her conclusion if she did not.

Model Answer — 6(b)

a preparation containing more enzyme provides more active sites, so more enzyme–substrate complexes form each second and the reaction is faster [1]

so a difference in rate could be caused by concentration rather than by any difference between the enzymes themselves [1]

the two variables would be confounded, and she could not conclude which enzyme is the better catalyst [1]

⚠ If you missed marks here: Confounding is the single idea being tested here, and the phrase “a difference in the result could have been caused by either variable” is worth having ready. Answers that stop at “it would not be a fair test” state the problem without explaining it.

1 mark — more enzyme gives more active sites and more complexes per second 1 mark — a difference in rate could then be caused by concentration, not by the enzymes 1 mark — the variables would be confounded, so no valid conclusion could be drawn

Preparation X clears the suspension in 40 s and preparation Y in 100 s under identical conditions. Calculate the rate for each using 1000 ÷ time, state which is faster and by how many times, and state one further piece of information the company would need before choosing.

Model Answer — 6(c)

X: 1000 ÷ 40 = 25 s⁻¹; Y: 1000 ÷ 100 = 10 s⁻¹ [1]

X is 2.5 times faster than Y at 30 °C and pH 7 [1]

further information needed: the cost of each preparation, or how each behaves over the whole range of temperature and pH used in the process, or how long each remains active (accept any sensible, justified suggestion) [1]

⚠ If you missed marks here: Both rates are needed for the first mark, and the comparison must be a ratio rather than a subtraction. The final mark rewards remembering that a single measurement at one temperature is a very narrow basis for a commercial decision.

1 mark — both rates calculated correctly as 25 and 10 s⁻¹ 1 mark — X identified as 2.5 times faster, with conditions stated 1 mark — a sensible additional piece of information suggested

(d) [2]

The technician writes: “Preparation X is the better enzyme.” Suggest why this conclusion is too broad.

Model Answer — 6(d)

the comparison was made at only one temperature and one pH, so it shows only that X is faster under those conditions [1]

Y might have a different optimum and be the faster enzyme at another temperature or pH, so the curves could cross — a wider comparison is needed before calling either one better [1]

⚠ If you missed marks here: This is the crossing-curves idea applied to a decision rather than a graph. A conclusion should always be limited to the conditions actually tested, and adding the phrase “under these conditions” is often the difference between a full-mark answer and an overstated one.

1 mark — recognition that only one temperature and pH were tested 1 mark — suggestion that Y might be faster under other conditions, so the curves could cross

Question 7 — Why Life Cannot Simply Turn Up the Heat

Total: 10 marks

A student calculates that a particular reaction of metabolism, uncatalysed, would take about 78 years to reach completion at 37 °C. The same reaction, catalysed by its enzyme, is complete in under a second. She suggests that an organism could avoid needing enzymes altogether by simply running at a much higher body temperature.

(a) [3]

Explain, using the figures given, why enzymes are described as necessary to sustain life.

Model Answer — 7(a)

uncatalysed, the reaction is far too slow — 78 years compared with under a second [1]

a cell must obtain energy and materials as fast as it uses them, so reactions occurring over years could not support any of the characteristics of living organisms [1]

enzymes increase the rate of every metabolic reaction so that it occurs fast enough at body temperature to sustain life [1]

⚠ If you missed marks here: The syllabus phrase “a reaction rate necessary to sustain life” is what the third mark is looking for. Use the figures rather than paraphrasing them — “much slower” is far weaker evidence than the comparison the question has handed you.

1 mark — comparison of the two times quoted from the stem 1 mark — a cell must supply energy and materials as fast as it spends them 1 mark — enzymes give a rate fast enough at body temperature to sustain life

(b) [4]

Explain why the student’s suggestion would not work.

Model Answer — 7(b)

an organism is built from proteins, including its enzymes and much of its structure [1]

at a high temperature those proteins would be denatured — the folded chains would unfold [1]

the shape of every active site would change so substrates would no longer be complementary to them [1]

so the organism would destroy the very molecules it depends on, and the damage would be permanent — catalysis is the only route available [1]

⚠ If you missed marks here: The question is really asking whether you understand why the two ways of speeding a reaction are not interchangeable for a living thing. Answers that stop at “it would be too hot” describe discomfort rather than biology; the mechanism must go through denaturation and the active site.

1 mark — organisms are built from proteins, including their enzymes 1 mark — high temperature would denature those proteins 1 mark — active sites would change shape so substrates no longer fit 1 mark — damage is permanent, so catalysis is the only available route

Some bacteria do live at 80 °C. Explain how this is possible without contradicting your answer to (b), and state one way in which such an organism is at a disadvantage.

Model Answer — 7(c)

their enzymes and other proteins are held in shape by stronger bonding, so they are not denatured at temperatures that would destroy human proteins [1]

their rate–temperature curves have the same shape but are shifted along the axis, with optima above 70 °C [1]

disadvantage: at ordinary temperatures such as 20 °C their enzymes are far below their optimum, with little kinetic energy and few effective collisions, so they grow very slowly or not at all [1]

⚠ If you missed marks here: The reconciliation is that the rule has not changed, only the temperature at which it applies. The third mark rewards seeing that a shifted curve moves both ends — heat tolerance is bought at the price of being useless in the cold.

1 mark — their proteins are held in shape by stronger bonding, so are not denatured at 80 °C 1 mark — the curve has the same shape but is shifted to higher temperatures 1 mark — disadvantage stated: very slow at ordinary temperatures because of low kinetic energy

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