Biology 9700/31 — May/June 2010
Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme
Topics Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation · Use of the Light Microscope
The enzyme E catalyses the hydrolysis of sucrose to produce fructose and glucose.
You are required to investigate the progress of this enzyme-catalysed reaction by finding the time taken for the decolourization of potassium permanganate.
The products of the hydrolysis of sucrose will change the colour of potassium permanganate as follows:
purple colourless
Test-tube Z shows the colourless end-point.
To follow the time course of this reaction take samples from the reaction mixture and test them with potassium permanganate.
Decide how often you will take these samples. You should not sample for longer than 20 minutes.
Write your sampling times in the space below.
Answer
0, 4, 8, 12, 16 (minutes)
Any list of 4 or more evenly-spaced times whose longest value is between 10 and 20 minutes scores full marks.
e.g. 0, 4, 8, 12, 16 (minutes)
Background Concept
In a time-course enzyme experiment the candidate samples the reaction mixture at known times to track how the rate changes. Sucrose is hydrolysed by the enzyme into fructose and glucose, both of which are reducing sugars that decolourise purple potassium permanganate. The faster the end-point is reached, the more reducing sugar was in the sample, so the time taken to decolourise the KMnO4 is an indirect measure of how much product has accumulated.
Understanding the Question
The candidate has to write down a list of times at which 5 cm³ samples will be removed from the beaker of reaction mixture. The whole experiment must finish within 20 minutes.
Approach
A typical CIE answer spaces the samples evenly (e.g. every 2, 3 or 4 minutes) and chooses the longest time so that the reaction has visibly progressed but the experiment is still practical to perform.
Step-by-Step Reasoning
- The mark scheme requires 4 or more sampling times.
- The times must be at even intervals (a constant gap between consecutive values).
- The longest time must be at least 10 minutes — long enough for the reaction to produce a measurable amount of reducing sugar.
- The longest time must be no more than 20 minutes because the question forbids sampling for longer than that.
- A safe choice is 0, 4, 8, 12, 16 minutes — five evenly-spaced times, last value 16 min, all within 20 min. Including time 0 lets the candidate capture the initial rate.
Key Takeaways
- Pick an evenly-spaced set of at least four sampling times.
- The last sample should be at or after 10 min and no later than 20 min.
- Include time 0 (the moment enzyme is added) so the initial rate can be seen.
Common Mistakes
- Only 3 times — the mark scheme requires 4 or more.
- Uneven spacing (e.g. 0, 2, 5, 11, 18) — rejected.
- Longest time < 10 min (e.g. 0, 2, 4, 6, 8) — the reaction has hardly progressed.
- Longest time > 20 min — outside the rule given in the question.
Things to Be Careful About
- Units are not required in this answer (and are ignored if given).
- Do not confuse these sampling times with the table to be drawn in (a)(ii); here you only list the times you intend to use.
Prepare the space below to record
- the time you remove each sample and
- the time at which the end-point is reached and
- the time taken to reach the end-point.
Answer
| Sampling time / min | Time at end-point / min:s | Time to reach end-point / s |
|---|---|---|
| 0 | 0:48 | 48 |
| 4 | 4:32 | 32 |
| 8 | 8:20 | 20 |
| 12 | 12:11 | 11 |
| 16 | 16:05 | 5 |
Key features that earn the marks:
- All cells drawn, heading line included, no outer boundary.
- Each column has a heading with a quantity and a unit (e.g. sampling time / min).
- Units are not repeated in the body of the table.
- The time to reach the end-point is given to the nearest whole second.
- The trend is correct: the time taken decreases (48 → 5 s) as the sampling time increases, because more reducing sugar is present later in the reaction.
Three-column table: Sampling time / min, Time at end-point / min:s, Time to reach end-point / s; whole-second values; decreasing trend.
Background Concept
The candidate has to design a table BEFORE doing the experiment. The table must hold three pieces of information per sample: when the sample was taken, when the end-point was reached, and how long the end-point took to appear. The biology then predicts what the recorded numbers should look like: as the reaction proceeds, more reducing sugar is present in the reaction mixture, so the potassium permanganate is decolourised more quickly. The time taken to reach the end-point therefore falls as the sampling time increases.
Understanding the Question
The candidate is given space in the answer booklet to draw a table. The table must be ready to record data BEFORE the experiment is carried out, and the expected trend should be visible from the layout (decreasing time taken as the reaction proceeds).
Approach
Follow the CIE Paper 3 table conventions: ruled cells, no outer box, headings in the top row (or left column) with the quantity AND the unit, units NOT repeated in the body, and a column for each piece of information. Then anticipate the trend: the time taken to reach the end-point decreases as the sampling time increases.
Step-by-Step Reasoning
- Mark 1 — table drawn with cells and a sampling-time column. All cells ruled, the heading is in the top row, and the first column is labelled with sampling time (or "time removed").
- Mark 2 — a second column heading with time and a unit. e.g. time at which end-point reached / min:s.
- Mark 3 — a third column with units clearly in the heading. e.g. time taken to reach end-point / s — units must be in the heading, not the body.
- Mark 4 — correct trend. Because more reducing sugar is present later in the reaction, the time taken to decolourise the KMnO4 should be shorter for later samples, so the number in the last row of the time-taken column is less than the number in the first row.
- Mark 5 — whole seconds or whole minutes for at least three results. The time to end-point is measured with a stop-clock, so values are whole seconds.
Key Takeaways
- A correct CIE results table has cells drawn, no outer box, headings with quantity + unit, and units not repeated in the body.
- The biology of the experiment dictates the trend: the time to decolourise KMnO4 decreases as the reaction proceeds.
- Time at the end-point is an elapsed (absolute) time from the start; time to end-point is a duration measured for the individual sample.
Common Mistakes
- Putting units in the body of the table (e.g. writing "45 s" rather than putting "s" in the heading).
- Omitting a column heading or putting it in the wrong place.
- Drawing an outer box around the whole table — the mark scheme says this is not required.
- Reversing the trend (time taken increasing) — biologically wrong and loses the trend mark.
- Writing times with decimals (e.g. 32.5 s) — the mark scheme requires whole seconds (or whole minutes).
Things to Be Careful About
- The two times are related: time-at-end-point = sampling time + time-taken, but they are recorded in separate columns.
- The trend mark can only be earned if the table contains at least two numbers in the time-taken column and the second is smaller than the first.
You are provided with
- 1% enzyme solution, labelled E
- 10% sucrose solution, labelled S
- sulfuric acid, labelled A
- 0.01% potassium permanganate solution, labelled P.
The sulfuric acid and potassium permanganate are harmful. If any comes into contact with your skin wash immediately under cold water. It is recommended that you wear safety goggles/glasses.
Proceed as follows:
- Label the test-tubes with the sampling times that you have decided to use.
- Put of A into each test-tube.
- Put of P into each test-tube. Gently shake each test-tube.
- Put of S into a beaker.
- Put of E into the beaker containing S.
- Immediately stir the reaction mixture in the beaker and start timing.
- At each of your sampling times, remove of the reaction mixture and add to the appropriate test-tube, mixing well.
- Immediately start timing and record the time taken to reach the end-point.
Describe a suitable control for this investigation.
Answer
Replace 4 cm³ of the enzyme solution E with 4 cm³ of boiled (and then cooled) enzyme solution — the denatured enzyme cannot catalyse the hydrolysis of sucrose, so any change in the reaction mixture is not due to enzyme activity.
(Equivalent alternative: replace 4 cm³ of E with 4 cm³ of water — no enzyme is present.)
Use boiled and cooled enzyme (or replace the enzyme with the same volume of water).
Background Concept
A control in an experiment is a tube (or beaker) treated identically to the test except for the variable being investigated. Its purpose is to show that any change observed in the test is actually caused by that variable. In an enzyme experiment the variable is the active enzyme itself, so a control must lack active enzyme.
Understanding the Question
The candidate has to describe a single control tube or beaker for the hydrolysis of sucrose by enzyme E.
Approach
There are two equivalent ways to remove the enzyme activity:
- Denature the enzyme by boiling (high temperature disrupts the tertiary structure of the protein and destroys the active site).
- Omit the enzyme and replace its volume with water.
Either of these is accepted by the mark scheme.
Step-by-Step Reasoning
- Boiling the enzyme irreversibly denatures it, so the reaction cannot proceed. The control must be allowed to cool before being added to the sucrose, otherwise the high temperature alone would introduce a second variable.
- Replacing the enzyme with water keeps the total volume (and therefore the concentration of sucrose) identical to the test, so the only difference is the presence/absence of active enzyme.
- Both controls show whether the observed decolourisation of KMnO4 in the test is due to enzyme activity, or whether the sucrose itself, the acid or the KMnO4 produce a change on their own.
Key Takeaways
- A control must differ from the test in only one variable — the one being investigated.
- For an enzyme investigation the standard controls are boiled (and cooled) enzyme, or enzyme replaced by water.
Common Mistakes
- Boiling the enzyme but adding it while still hot — this introduces a second variable (temperature).
- Replacing enzyme with a different buffer or with acid — this changes more than one variable.
- Saying "do the experiment without the substrate" — the question is about the enzyme, so sucrose should still be present.
Things to Be Careful About
- A control is not a "repeat". A repeat is a duplicate of the test; a control has the active ingredient removed or denatured.
Identify two significant sources of error in this investigation.
Answer
Any two of:
- Timing is imprecise — it is hard to remove the 5 cm³ sample at the exact moment stated and simultaneously start the second stop-clock for the end-point.
- Judging the end-point is subjective — it is difficult to decide exactly when the purple colour has gone colourless.
- The volume of the reaction mixture decreases as each 5 cm³ sample is removed, so the concentration of sucrose (and later of products) changes during the experiment.
Two of: imprecision in timing; subjectivity of the colour end-point; decreasing volume of the reaction mixture.
Background Concept
A source of error is a step in the procedure that introduces variability into the measurements. Significant sources of error are those that have a real effect on the recorded data and cannot be eliminated by a careful candidate working alone.
Understanding the Question
The candidate has to identify two distinct, significant sources of error in the procedure described in the question (steps 1–8).
Approach
The mark scheme lists three valid sources of error. Read the procedure carefully and pick the two that are the most serious in this particular experiment.
Step-by-Step Reasoning
- Timing error. There are TWO separate timings: when the sample is removed from the reaction beaker, and when the KMnO4 is decolourised. Starting and stopping two stop-clocks while pipetting a 5 cm³ sample into a test-tube is hard to do exactly. The procedure compounds the error because any slip in the first timing carries into the second.
- Subjective end-point. "Colourless" is a judgement, not a discrete event. Different candidates (or the same candidate on different occasions) will declare the end-point at slightly different shades of pink. This is especially true as the reaction slows down and the colour change takes longer to complete.
- Decreasing volume. Each sample removes 5 cm³ from the beaker. After five samples only 9 cm³ of the original 34 cm³ is left, so the concentrations of sucrose, enzyme and water have all changed appreciably. This is NOT a property of the test-tubes themselves; it is an error in the assumption that the reaction mixture is unchanged throughout the experiment.
Key Takeaways
- Significant sources of error are real, procedure-specific difficulties — not vague generalities like "human error" or "parallax".
- Volume changes in the reaction vessel are an easy error to overlook in a time-course experiment.
- The KMnO4 end-point is subjective and is a common source of error in any colorimetric practical.
Common Mistakes
- Naming temperature or pH as an error — the mark scheme explicitly rejects these (they are covered in c(ii) as uncontrolled variables).
- Naming evaporation — also rejected.
- Saying "human error" or "not accurate enough" without saying WHAT is hard to do.
- Confusing sources of error with limitations/improvements (c(ii)).
Things to Be Careful About
- "Error" here means a cause of variability or inaccuracy, not a mistake. The mark scheme rejects temperature, pH and evaporation as sources of error in this experiment.
State one variable which was not controlled in this investigation and how it could have been controlled.
Answer
Temperature of the reaction mixture was not controlled. → Place the beaker of reaction mixture in a thermostatically-controlled water-bath (or a water-bath at a set temperature), e.g. 25 °C, throughout the experiment.
(Equivalent alternative: pH was not controlled. → Use a buffer of constant pH in place of, or in addition to, the sulfuric acid.)
Temperature was uncontrolled — use a thermostatically-controlled water-bath.
Background Concept
Enzyme activity is strongly affected by temperature and pH. If either of these varies during the experiment, the measured rate will vary for reasons that have nothing to do with the independent variable (time / substrate concentration). A good procedure therefore holds them constant.
Understanding the Question
The candidate has to identify ONE variable that is not held constant in this procedure and say how it could be controlled.
Approach
Look at the procedure: the beaker of reaction mixture is not placed in any temperature-regulating device, and the only pH-controlling reagent is a fixed amount of sulfuric acid. Either temperature or pH (mark scheme accepts either) can be named, with a corresponding practical method of control.
Step-by-Step Reasoning
- Temperature. The reaction rate roughly doubles for every 10 °C rise. A bench at "room temperature" can easily vary by 2–3 °C during a 20-minute experiment, and a beaker sitting in a sunny spot can warm up further. The fix is to stand the beaker in a thermostatically-controlled water-bath (or a large beaker of water at a measured temperature, topped up if necessary).
- pH. The procedure adds 1 mol dm⁻³ sulfuric acid to the test-tubes but no buffer to the reaction beaker. As sucrose is hydrolysed, hydrogen ions may be released or consumed, and the small amount of acid in the test-tubes does not control the pH of the bulk reaction. A buffer of constant pH added to the reaction beaker would control this.
Key Takeaways
- Temperature and pH are the two variables most commonly left uncontrolled in school enzyme experiments.
- A control method must be practical in a school lab — a thermostatically-controlled water-bath is a standard piece of equipment.
Common Mistakes
- Naming more than one variable — the mark scheme says "max 1".
- Giving an improvement without naming the variable (or vice versa).
- Suggesting "use a thermometer" — measuring is not the same as controlling.
Things to Be Careful About
- "Use a buffer" is the right control for pH, not "add more acid".
- "Water-bath at constant temperature" needs a thermostat, or at least a thermometer to monitor it.
Table 1.1 shows the results for a similar investigation which measured the mass of reducing sugars produced over a period of 400 seconds.
Table 1.1
| time / s | mass of reducing sugars / mg |
|---|---|
| 60 | 0.32 |
| 120 | 0.64 |
| 180 | 0.95 |
| 300 | 1.55 |
| 400 | 2.05 |
Plot a graph of the data shown in Table 1.1.
Answer
Plotted graph on the printed grid (Fig. 1.1):
- x-axis: time / s, scale 100 s to 2 cm, range 0 – 400 s.
- y-axis: mass of reducing sugars / mg, scale 0.5 mg to 2 cm, range 0 – 2.5 mg.
- Plotted points (small ×) at: (60, 0.32), (120, 0.64), (180, 0.95), (300, 1.55), (400, 2.05).
- Single straight line of best fit drawn through the points, extrapolated back to the origin (0, 0).
Plotted points at (60, 0.32), (120, 0.64), (180, 0.95), (300, 1.55), (400, 2.05) with a straight line of best fit extrapolated to the origin.
Background Concept
A line graph is the correct way to display two continuous variables (time and mass) where one is being measured at a series of values of the other. A bar chart or histogram would be wrong here because the data are not categorical.
Understanding the Question
The candidate has to plot the five pairs of values from Table 1.1 on the printed grid (Fig. 1.1) and draw the line of best fit.
Approach
Use the four CIE marks for graph drawing in order: axes and labels (O), scale (S), plotting (P), line (L).
Step-by-Step Reasoning
- O — axes and labels. x-axis is time with unit s (or sec/seconds); y-axis is mass of (reducing) sugars with unit mg. Both labels must be present and the units must be written.
- S — scale. The mark scheme requires 100 s to 2 cm on the x-axis and 0.5 mg to 2 cm on the y-axis. Awkward scales (e.g. 3 s to 1 cm, or 0.7 mg to 1 cm) lose the scale mark even if the points are plotted correctly.
- P — plotting. Use small crosses (×) or dots in a circle (⊙). The intersection of the cross must be clear. Blobs, large filled circles and plain dots are rejected.
- L — line. A single, thin, continuous straight line of best fit through the points. In this case the data fall almost exactly on a straight line through the origin, so the line should be extrapolated back to (0, 0). Do NOT join the points plot-to-plot with a zig-zag, and do not extrapolate the line beyond the axes.
Key Takeaways
- A correct CIE graph has labels with units, a sensible scale (no awkward numbers), clearly-plotted crosses, and a single straight or smoothly-curved best-fit line.
- For a reaction whose rate is constant, the points lie on (or close to) a straight line through the origin.
Common Mistakes
- Missing units on the axis labels.
- Using awkward scales (e.g. 30 s to 1 cm, or 0.3 mg to 1 cm).
- Plotting with filled blobs or by writing the number next to a dot.
- Joining the points plot-to-plot instead of drawing a best-fit line.
- Extrapolating the line beyond the printed axes.
Things to Be Careful About
- The grid given is wider than it is tall, so the natural scales are 100 s to 2 cm on x and 0.5 mg to 2 cm on y. Use these unless the candidate is using an ECFs scale for a special reason.
- Re-check the plotted points against the table before drawing the line — a mis-plotted point drags the line off centre.
Use your graph to find the rate of hydrolysis of the sucrose by finding the gradient of the line.
Show on your graph where you took the readings to calculate the gradient.
Show all the steps in your calculation.
rate of enzyme activity = ______
Working
Mark two well-separated points on the line of best fit and read off their (time, mass) coordinates. Using the extrapolated line, the simplest pair is the origin (0 s, 0 mg) and the top of the data range (400 s, 2.05 mg):
(Using the data-table endpoints (60 s, 0.32 mg) and (400 s, 2.05 mg) gives a very similar answer:
— the mark scheme credits the simpler 2.05 / 400 form.)
Answer
(≈ 0.00513 mg s⁻¹)
5.13 × 10⁻³ mg s⁻¹ (≈ 0.00513 mg s⁻¹)
Background Concept
The gradient of a line on a graph of mass-of-product (y) against time (x) is the rate of product formation. For a reaction whose rate is constant over the time interval studied, the line is straight and the gradient is the same everywhere on the line. The gradient is calculated as using two well-separated points so that any reading error is a small fraction of the difference.
Understanding the Question
The candidate has to find the rate of the enzyme-catalysed hydrolysis by calculating the gradient of the line drawn in (d)(i). The readings used must be marked on the graph, the calculation shown, and the answer written with the unit .
Approach
Draw a large gradient triangle on the line of best fit, with the two right-angle corners on two well-separated points of the line. Read off the two (time, mass) values, divide the difference in mass by the difference in time, and round the final answer to 3 significant figures.
Step-by-Step Reasoning
- Mark 1 — at least one time and one mass shown on the graph. Draw a small triangle or simply write the two chosen points on the line.
- Mark 2 — two masses AND two times clearly shown. e.g. 2.05 mg at 400 s and 0 mg (or 0.32 mg) at 0 s (or 60 s).
- Mark 3 — the calculation explicitly shows 2.05 mg ÷ 400 s (or equivalent). The mark scheme credits the use of 2.05 mg and 400 s as the upper values.
- Mark 4 — final answer rounded to ≤ 3 significant figures (or 5 decimal places, or standard form). 2.05 / 400 = 0.005125. Three sig figs gives 0.00513; standard form gives .
- The unit must be written: .
Key Takeaways
- A gradient is calculated from two well-separated points on the line, not from the raw data table.
- The rate of a product-forming reaction is the gradient of a mass-vs-time graph, in mass per unit time.
- The mark scheme credits the maximum values (2.05 mg and 400 s) — the candidate does not have to use exactly these, but the calculation should reach the same order of magnitude.
Common Mistakes
- Dividing mass by mass (or time by time) instead of mass by time.
- Using only one pair of values (e.g. 0.32 ÷ 60) — this is the rate between two data points, not the gradient of the best-fit line.
- Forgetting the unit .
- Reporting 0.00513 as 0.0051 (only 2 sig figs) — the mark scheme wants at most 3 sig figs, not "at most 2".
- Reading the coordinates off the data table rather than off the line of best fit.
Things to Be Careful About
- The mark scheme gives credit for "any answer rounded to maximum of three significant figures". 0.00513 IS three sig figs (the leading zeros do not count). 0.0051 is only two.
- The mark scheme also accepts five decimal places (0.00513) and standard form ().
- The candidate should always show the working on the graph (the triangle) AND as a calculation below it.
The student repeated the investigation for 600 seconds and found that the results for the enzyme-catalysed reaction between 60 and 400 seconds were similar to the results in Table 1.1 but after 400 seconds the mass of reducing sugars produced remained the same.
Explain why the mass of reducing sugars increased and then remained the same.
Answer
- (Increase, 0 – 400 s) The enzyme catalyses the hydrolysis of the non-reducing sugar (sucrose) into reducing sugars (glucose and fructose), so the mass of reducing sugars rises.
- (Plateau, after 400 s) All of the substrate has been hydrolysed (broken down / used up), so no further reducing sugars can be produced and the mass remains constant.
[Reject: "enzyme active sites full" or "enzyme used up".]
Enzyme hydrolyses sucrose into reducing sugars; after 400 s all the substrate is used up so the mass stays constant.
Background Concept
Enzyme-catalysed reactions have a finite substrate supply. As long as substrate molecules are available to bind to the active site, product is made at a roughly constant rate. Once the substrate is exhausted the reaction stops, even though the enzyme itself is still active. The graph of product mass against time therefore rises (often linearly at first) and then flattens into a plateau.
Understanding the Question
The student in the question is told that the data between 60 and 400 s are similar to Table 1.1, but that the mass of reducing sugars stayed the same after 400 s. The candidate has to explain this pattern in terms of the underlying biology.
Approach
Two marks are available: one for explaining the increase, one for explaining the plateau. The mark scheme ties the increase to the enzyme hydrolysing sucrose and the plateau to the substrate running out.
Step-by-Step Reasoning
- Mark 1 — enzyme and substrate. The enzyme E is catalysing the hydrolysis of the non-reducing sugar (sucrose) to give the reducing sugars glucose and fructose. The mass of reducing sugars therefore increases with time. The mark scheme explicitly REQUIRES the word "enzyme" here.
- Mark 2 — substrate used up. After 400 s the rate of formation of reducing sugar drops to zero, because all of the sucrose has been hydrolysed. There is no more substrate for the enzyme to act on, so the mass cannot increase further. The mark scheme accepts "all substrate hydrolysed / broken down / used up".
- What is NOT accepted. "The enzyme's active sites are full" or "the enzyme is used up". These are biologically wrong: there is plenty of free enzyme, but no substrate to bind to it. The candidate must phrase the explanation in terms of the substrate, not the enzyme, running out.
Key Takeaways
- A product-against-time graph that rises linearly and then plateaus is the classic signature of a reaction limited by substrate availability.
- In the explanation, "substrate" or "non-reducing sugar" must be named — generic statements about "the reaction stopping" do not earn the mark.
Common Mistakes
- Saying "the enzyme has been used up" or "denatured" — the enzyme is unchanged; the substrate has gone.
- Saying "the active sites are full" — this implies an excess of substrate, which is the opposite of what is happening.
- Confusing reducing with non-reducing sugars — sucrose is non-reducing; glucose and fructose are reducing.
- Saying "the reaction has reached equilibrium" — the reaction goes to completion (all the sucrose is hydrolysed); it is not a reversible equilibrium under these conditions.
Things to Be Careful About
- The mark scheme says "Reject enzyme active sites full or enzyme used up". Either of these will cost the candidate the substrate-depletion mark even if the enzyme point is correct.
The rest of this paper
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